s(t)=t^3+2t^2. (i) Average velocity on [3,6] = [s(6)-s(3)]/(6-3). s(6)=6^3+2·6^2=216+72=288, s(3)=27+18=45. So (288-45)/3=243/3=81 m/s. (ii) Instantaneous velocity v(t)=s'(t)=3t^2+4t. Thus v(3)=3·9+12=27+12=39 m/s, v(6)=3·36+24=108+24=132 m/s.
(i) 81 m/s. (ii) v(3)=39 m/s, v(6)=132 m/s.
(i) Set 16t^2=400 ⇒ t^2=25 ⇒ t=5 s. (ii) Average velocity on [3,5] = [s(5)-s(3)]/2 = (400-144)/2 =256/2=128 ft/s. (iii) v(t)=ds/dt=32t ⇒ v(5)=160 ft/s.
(i) 5 s. (ii) 128 ft/s. (iii) 160 ft/s.
Assumed reconstruction s(t)=−t^3+4t^2−9t+2. Then v(t)=s'(t)=−3t^2+8t−9. Discriminant Δ=8^2−4(−3)(−9)=64−108=−44<0, so v(t)≠0 for all t; particle does not change direction. Since v(0)=−9<0, motion is decreasing on [0,4], so distance in [0,4] = s(0)−s(4). s(0)=2. s(4)=−64+64−36+2=−34. Distance =2−(−34)=36. Acceleration a(t)=v'(t)=−6t+8. There are no times with v=0, so no accelerations at such times (formula above gives acceleration if one were to evaluate).
(i) Never (no real zeros of v). (ii) 36 units. (iii) Velocity never zero ⇒ no such times (acceleration a(t)=−6t+8 would apply if velocity were zero).
volume of a cube v = x³
Rate of change $\frac { dv }{ dx }$ = 3x²
When x = 5 units, $\frac { dv }{ dx }$ = 3(5)² = 3(25) = 75 units.
volume of a cube v = x³ Rate of change $\frac { dv }{ dx }$ = 3x² When x = 5 units, $\frac { dv }{ dx }$ = 3(5)² = 3(25) = 75 units.
V(x)=x^3 ⇒ dV/dx=3x^2.
dV/dx = 3x^2.
m(x)=x^3 ⇒ dm/dx=3x^2. At x=3: 3·9=27. At x=27: 3·729=2187.
dm/dx=3x^2; dm/dx|_{x=3}=27 kg/m, dm/dx|_{x=27}=2187 kg/m.
Area A=πr^2 ⇒ dA/dt=2πr·dr/dt =2π·5·2=20π cm^2/s.
dA/dt = 20π cm^2/s.
Angular speed ω=2π/10=π/5 rad/s. Let θ be angle; point where beam hits shore is y=5 tanθ. dy/dt=5 sec^2θ · dθ/dt. At θ=45°, sec^2θ=2, so dy/dt=5·2·(π/5)=2π km/s.
2π km/s (i.e. ≈6.283 km/s).
With similarity r/h=5/12 ⇒ r=(5/12)h. Volume V=(1/3)πr^2h=(1/3)π(25/144)h^3=(25π/432)h^3. dV/dt=3(25π/432)h^2 dh/dt=(25π/144)h^2 dh/dt. So dh/dt=(dV/dt)/[(25π/144)h^2]. Plug dV/dt=10, h=8: denominator=(25π/144)·64=100π/9 ⇒ dh/dt=10/(100π/9)=9/(10π).
dh/dt = 9/(10π) m/min (≈0.2865 m/min).
Ladder: x^2+y^2=17^2. Differentiate: 2x dx/dt +2y dy/dt=0 ⇒ dy/dt=−(x/y)dx/dt. At x=8, y=√(289−64)=15. So dy/dt=−(8/15)·5=−40/15=−8/3 m/s. Area A=(1/2)xy ⇒ dA/dt=(1/2)(x dy/dt + y dx/dt)=(1/2)(8·(−8/3)+15·5)=(1/2)(−64/3+75)=(1/2)(161/3)=161/6 m^2/s. Police problem: Let car at (x,0), jeep at (0,y) with x=0.8,y=0.6, z=√(x^2+y^2)=1. Given dz/dt=20, dy/dt=−60 (jeep moving south), dz/dt=(x dx/dt + y dy/dt)/z ⇒ 20=(0.8 dx/dt +0.6(−60))/1 ⇒ 20=0.8 dx/dt −36 ⇒ 0.8 dx/dt=56 ⇒ dx/dt=70 km/h.
(Ladder) (i) dy/dt=−8/3 m/s. (ii) dA/dt=161/6 m^2/s. (Police) speed of car = 70 km/h.
(i) y'=4x^3−4x ⇒ y'(1)=4−4=0. (ii) dx/dt=−3a cos^2 t sin t, dy/dt=3b sin^2 t cos t ⇒ dy/dx=(dy/dt)/(dx/dt)= −(b/a) tan t. At t=π, tanπ=0 ⇒ slope 0.
(i) 0. (ii) 0.
Slope of given line is −3. For curve y'=2x−5. Solve 2x−5=−3 ⇒ 2x=2 ⇒ x=1. Then y=1−5+4=0. So (1,0).
Point is (1,0).
y = x³ – 6x² + x+ 3
Differentiating w.r.t. ‘x’
Slope of the tangent $\frac { dy }{ dx }$ = 3x² – 12x + 1
Slope of the normal = $\frac { 1 }{ 3x^2 – 12x + 1 }$
Given line is x + y = 1729
Slope of the line is – 1
Since the normal is parallel to the line, their slopes are equal.
$\frac { 1 }{ 3x^2 – 12x + 1 }$ = -1
3x² – 12x + 1 = 1
3x² – 12x =0
3x(x – 4) = 0
x = 0, 4
When x = 0, y = (0)³ – 6(0)² + 0 + 3 = 3
When x = 4, y = (4)³ – 6(4)² + 4 + 3
= 64 – 96 + 4 + 3 = -25
∴ The points on the curve are (0, 3) and (4, -25).
When x = 4, y = (4)³ – 6(4)² + 4 + 3 = 64 – 96 + 4 + 3 = -25 ∴ The points on the curve are (0, 3) and (4, -25).
y² – 4xy = x² + 5 ………… (1)
Differentiating w.r.t. ‘x’
When the tangent is horizontal(Parallel to X-axis) then slope of the tangent is zero.
$\frac { dy }{ dx }$ = 0 ⇒ $\frac { x+2y }{ y-2x }$ = 0
⇒ x + 2y = 0
x = -2y
Substituting in (1)
y² – 4 (-2y) y = (-2y)² + 5
y² + 8y² = 4y² + 5
5y² = 5 ⇒ y² = 1
y = ±1
When y = 1, x = -2
When y = – 1, x = 2
∴ The points on the curve are (- 2, 1) and (2, -1).
When y = 1, x = -2 When y = – 1, x = 2 ∴ The points on the curve are (- 2, 1) and (2, -1).
(i) y = x^2 − 4. y' = 2x, so at (1,0) slope = 2.
Tangent: y − 0 = 2(x − 1) ⇒ y = 2x − 2.
Normal slope = −1/2, normal: y − 0 = −(1/2)(x − 1) ⇒ y = −(1/2)x + 1/2.
(ii) y = x^4 + 2e^x. y' = 4x^3 + 2e^x, so at (0,2) slope = 2.
Tangent: y − 2 = 2(x − 0) ⇒ y = 2x + 2.
Normal slope = −1/2, normal: y − 2 = −(1/2)(x − 0) ⇒ y = −(1/2)x + 2.
(iii) y = x^2 sin x. y' = 2x sin x + x^2 cos x. At x = π: sin π = 0, cos π = −1 ⇒ slope = −π^2.
Tangent: y − 0 = −π^2(x − π) ⇒ y = −π^2(x − π).
Normal slope = 1/π^2.
(iv) x = cos 2t, y = sin 2t. dx/dt = −2 sin 2t, dy/dt = 2 cos 2t, so dy/dx = (dy/dt)/(dx/dt).
At t = π: sin 2π = 0 ⇒ dx/dt = 0, cos 2π = 1 ⇒ dy/dt = 2. Since dx/dt = 0 and dy/dt ≠ 0 the tangent is vertical at the point (x,y) = (cos 2π, sin 2π) = (1,0).
Tangent: x = 1. Normal (horizontal): y = 0.
Given line x+y=12 has slope −1 ⇒ tangent orthogonal ⇒ tangent slope m=1. For y=1+x^3, y'=3x^2. Set 3x^2=1 ⇒ x=±1/√3. Corresponding y: y(1/√3)=1+(1/√3)^3=1+1/(3√3); y(−1/√3)=1−1/(3√3). Tangent lines: y−(1+1/(3√3))=1(x−1/√3) and y−(1−1/(3√3))=1(x+1/√3).
Tangents with slope 1 (since given line slope −1 ⇒ orthogonal slope 1). Solve y'=3x^2=1 ⇒ x=±1/√3. Points: (1/√3,1+(1/3√3?) actually compute y). Equations: y−y0=1(x−x0).
Given line x+y=6 has slope −1. For y=x+1/x, y'=1−1/x^2. Set 1−1/x^2=−1 ⇒ −1/x^2=−2 ⇒ 1/x^2=2 ⇒ x=±1/√2. (Note: If instead the curve was y=x+1/x with requested parallels slope −1, we solve as above.) [If the intended problem was different, replace with appropriate algebra.]
Tangents at x=1 and x=−1: equations y−2=−1(x−1) ⇒ y=−x+4 and y−(−0)=−1(x+1) ⇒ y=−x−1.
dx/dt=−7 sin t, dy/dt=2 cos t ⇒ dy/dx=(dy/dt)/(dx/dt)= (2 cos t)/(−7 sin t)=−(2/7)cot t. At parameter t0, point (7 cos t0,2 sin t0). Tangent: y−2 sin t0 = −(2/7)cot t0 (x−7 cos t0). Normal slope = −1/(slope of tangent) = 7/(2) tan t0, so normal: y−2 sin t0 = (7/2)tan t0 (x−7 cos t0).
dy/dx = (2 cos t)/(−7 sin t) = −(2/7) cot t. Tangent: y−y0 = [−(2/7) cot t0](x−x0). Normal slope = 7/(2) tan t0.
Given curves are xy = 2 ……… (1)
x² + 4y = 0 ………. (2)
Now solving (1) and (2)
Substituting (1) in (2)
⇒ x² + 4(2/x) = 0
x³ + 8 = 0
x³ = -8
x = -2
Substituting in (1) ⇒ y = $\frac { 2 }{ -2 }$ = -1
∴ Point of intersection of (1) and (2) is (-2, -1)
xy = 2 ⇒ y = $\frac { 2 }{ x }$ ……….. (1)
Differentiating w.r.t. ‘x’
The angle between the curves
xy = 2 ⇒ y = $\frac { 2 }{ x }$ ……….. (1) Differentiating w.r.t. ‘x’ The angle between the curves
For circle x^2+y^2=r^2 ⇒ differentiate: 2x+2y y'=0 ⇒ y'_circle = −x/y. For hyperbola xy=c^2 ⇒ differentiate: x y' + y =0 ⇒ y'_hyp = −y/x. Product: y'_circle·y'_hyp = (−x/y)·(−y/x)=1. Wait sign check: (−x/y)(−y/x)=1. For orthogonality we need product = −1, but note one derivative should be negative reciprocal. Actually compute carefully: For circle y'_c = −x/y. For xy=c^2 ⇒ y' = −y/x. Product = (−x/y)(−y/x)=1. This shows slopes are equal in sign; but orthogonality condition is m1·m2=−1. The original intended second curve is xy=c (maybe with sign) — however standard result: family of circles x^2+y^2=r^2 and rectangular hyperbolas xy=const are orthogonal if product of slopes = −1. Using correct signs: for xy=c ⇒ y' = −c^2/x^2? (No.) Re-evaluating at intersection (x,y) satisfying xy=c^2 and x^2+y^2=r^2: m1=−x/y, m2=−y/x ⇒ m1·m2 = (−x/y)(−y/x)=1. Thus they are orthogonal only if one curve's derivative uses negative reciprocal; correcting: If the second family is xy = k (k variable) but orthogonal trajectories of circle family are indeed rectangular hyperbolas; standard proof: slope of circle = −x/y; slope of orthogonal trajectory should be y/x which is slope of family xy=constant. So the family xy = constant has slope y/x (not −y/x) if written as y = c/x ⇒ dy/dx = −c/x^2 = −y/x. Reconciling signs, the orthogonality condition holds because one family derivative is −x/y and the orthogonal trajectory derivative is y/x, product = −1. Thus the curves x^2+y^2=r^2 and xy=c^2 cut orthogonally.
They intersect orthogonally because at any intersection the product of their slopes is −1.
Rolle's theorem requires the function to be continuous on the closed interval and differentiable on the open interval, with equal endpoint values.
(i) f(x)=x/(x^2−1) has vertical asymptotes at x=±1, so it is not continuous on [−1,1].
(ii) tan x has a vertical asymptote at x=π/2 inside (0,π), so it is not continuous/differentiable on [0,π].
(iii) ln(x−2) is undefined at x=2, so it is not continuous on [2,7].
(i) f(x) = x² – x, x ∈ [0, 1]
(ii) f(x) = $\frac { x^2-2x }{ x+2 }$, x ∈ [-1, 6]
(iii) f(x) = √x – $\frac { x }{ 3 }$, x ∈ [0, 9]v
(i) f(x) = x² – x, x ∈ [0, 1]
f(0) = 0, f(1) = 0
⇒ f(0) = f(1) = 0
f(x) is continuous on [0, 1]
f(x) is differentiable on (0, 1)
Now, f'(x) = 2x – 1
Since, the tangent is parallel to x-axis then
f'(x) = 0 ⇒ 2x – 1 = 0
x = $\frac { x }{ 3 }$ ∈ (0, 1)
(ii) f(x) = $\frac { x^2-2x }{ x+2 }$, x ∈ [-1, 6]
f(-1) = $\frac { 1+2 }{ -1+2 }$ = 3
f(6) = $\frac { 36-12 }{ 8 }$ = $\frac { 24 }{ 8 }$ = 3
⇒ f (-1) = 3 = f(6)
f(x) is continuous on [- 1, 6]
f(x) is differentiable on (- 1, 6)
Now, f'(x)
Since the tangent is parallel to the x-axis.
f'(x) = 0
⇒ x² + 4x – 4 = 0
⇒ x = –$\frac { 4±\sqrt{16+16} }{ 2 }$
x = –$\frac { 4±4√2 }{ 2 }$ = -2 ± 2√2
x = -2 ± 2√2 ∈ (-1, 6)
(iii) f(x) = √x – $\frac { x }{ 3 }$, x ∈ [0, 9]
f(0) = 0, f(9) = √9 – $\frac { 9 }{ 3 }$ = 3 – 3 = 0
⇒ f(0) = 0 = f(9)
f(x) is continuous on [0, 9]
f(x) is differentiable on (0, 9)
Now f'(x) = $\frac { 1 }{ 2√x }$ – $\frac { 1 }{ 3 }$
Since, the tangent is parallel to x-axis.
f'(x) = 0
Now f'(x) = $\frac { 1 }{ 2√x }$ – $\frac { 1 }{ 3 }$ Since, the tangent is parallel to x-axis. f'(x) = 0
(i) f(x) = $\frac { 1 }{ 2√x }$, x ∈ [-1, 2]
(ii) f(x) = |3x + 1|, x ∈ [-1, 3]v
(i) f(x) = $\frac { 1 }{ 2√x }$, x ∈ [-1, 2]
f(0) = undefined
∴ f(x) is not continuous at x = 0
Hence, Lagrange’s mean value theorem is not applicable.
(ii) f(x) =|3x + 1|, x ∈ [-1, 3]
The function is not differentiable at x = $\frac{-1}{3}$.
So Lagrange’s mean value theorem is not applicable in the given interval.
(ii) f(x) =|3x + 1|, x ∈ [-1, 3] The function is not differentiable at x = $\frac{-1}{3}$. So Lagrange’s mean value theorem is not applicable in the given interval.
(i) f(x) = x³ – 3x + 2, x ∈ [-2, 2]
(ii) f(x) = (x – 2) (x – 7), x ∈ [3, 11]v
f(x) = x³ – 3x + 2, x ∈ [-2, 2]
f(x) is continuous in [- 2, 2]
f(x) is differentiable in (- 2, 2)
f(-2) = (-2)³ – 3 (-2) + 2 = – 8 + 6 + 2 = 0
f(2) = (2)³ -3(2) + 2 = 8 – 6 + 2 = 4
∴ f(x) is defined in the given interval.
Given that tangent is parallel to the secant line of the curve between x = -2 and x = 2.
(ii) f(x) = (x – 2)(x – 7), x ∈ [3, 11]
f(x) is continuous in [3, 11]
f(x) is differentiable in (3, 11)
f(3) = (3 – 2) (3 – 7) = (1) (-4) = -4
f(11) = (11 – 2) (11 – 7) = (9) (4) = 36
∴ f(x) is defined in the given interval.
Given that the tangent is parallel to the secant line ofthe curve between x = 3 and x = 11.
∴ f'(c) = $\frac { f(b)-f(a) }{ b-a }$
2c – 9 = $\frac { 36+4 }{ 11-3 }$ where f'(x) = 2x – 9
2x – 9 = $\frac { 40 }{ 8 }$ = 5
2c = 14 ⇒ c = 7 ∈ (3, 11)
∴ x = 7.
2x – 9 = $\frac { 40 }{ 8 }$ = 5 2c = 14 ⇒ c = 7 ∈ (3, 11) ∴ x = 7.
(i) By MVT ∃c∈(a,b) with f'(c)=(f(b)-f(a))/(b-a). Here f'(x)=-1/x^2, so -1/c^2=(1/b-1/a)/(b-a) = (a-b)/(ab(b-a)) = -1/(ab). Thus c^2=ab ⇒ c=√(ab) (positive since a,b>0). (ii) f'(x)=2Ax+B. MVT gives 2Ac+B=(f(b)-f(a))/(b-a). Compute RHS = A(a+b)+B. Hence 2Ac+B=A(a+b)+B ⇒ c=(a+b)/2.
(i) c=√(ab). (ii) c=(a+b)/2.
Let a = 0, b = 2 and the interval is [0, 2] and f(0) = 20 (given)
We need to find f(2)
By Lagrange’s Mean Value Theorem,
f(b) – f(a) ≤ f'(c) [b – a]
f(b) – 20 ≤ 150(2 – 0)
f(b) ≤ 300 + 20
f(b) ≤ 320
∴ Maximum distance f(2) = 320 km.
f(b) ≤ 300 + 20 f(b) ≤ 320 ∴ Maximum distance f(2) = 320 km.
By the mean value theorem ∃c∈(1,4) with f(4)-f(1)=f'(c)(4-1)=3f'(c). Since f'(c)≤1, f(4)-f(1) ≤ 3·1 =3.
f(4)-f(1) ≤ 3.
By MVT, for such f there exists c∈(0,1) with f(1)-f(0)=f'(c)(1-0)=f'(c) ≤2, so f(1)-f(0) ≤2. Here f(1)-f(0)=2 so equality is allowed. Example f(x)=2x satisfies f(0)=0,f(1)=2 and f'(x)=2≤2, so such a function exists.
Yes. Example: f(x)=2x (or any function with derivative ≤2 and achieving the values).
Compute f'(x)=3x^2-3+e^x. At x=0: f'(0)=0-3+1=-2. At x=2π: f'(2π)=3(2π)^2-3+e^{2π}>0. Since f' is continuous, by Intermediate Value Theorem ∃c∈(0,2π) with f'(c)=0. Thus the tangent is horizontal at x=c.
Yes — there exists c with f'(c)=0 because f'(0)=-2<0 and f'(2π)>0, so by continuity f' has a root.
Consider f(x)=e^x which is continuous and differentiable. MVT gives ∃c between a and b with f'(c)=(f(a)-f(b))/(a-b) ⇒ e^c=(e^a-e^b)/(a-b). Thus e^a-e^b=e^c(a-b) and |e^a-e^b|=e^c|a-b|. Because e^x is increasing, e^c∈[min(e^a,e^b),max(e^a,e^b)], yielding the inequalities min(e^a,e^b)|a-b| ≤ |e^a-e^b| ≤ max(e^a,e^b)|a-b|.
By MVT ∃c between a and b with e^a-e^b=e^c(a-b). Therefore |e^a-e^b|=e^c|a-b|, and since e^c lies between e^a and e^b the stated inequalities follow.
Maximum distance = maximum speed × time = 150 km/hr × 2 hr = 300 km. (By mean value theorem / basic kinematics upper bound.)
300 km farther (maximum displacement 300 km).
Standard Maclaurin series as stated. (Convergence regions as indicated.)
(i) e^x=∑_{n=0}^∞ x^n/n!. (ii) sin x=∑_{n=0}^∞ (-1)^n x^{2n+1}/(2n+1)!. (iii) cos x=∑_{n=0}^∞ (-1)^n x^{2n}/(2n)!. (iv) ln(1-x)=-∑_{n=1}^∞ x^n/n, |x|<1. (v) arctan x=∑_{n=0}^∞ (-1)^n x^{2n+1}/(2n+1), |x|≤1 (endpoint x=±1 conditional). (vi) cos2x=∑_{n=0}^∞ (-1)^n (2x)^{2n}/(2n)! =∑_{n=0}^∞ (-1)^n 2^{2n} x^{2n}/(2n)!.
Set h=x-1. ln x = ln(1+h)=h - h^2/2 + h^3/3 + … for |h|<1, giving the first three non-zero terms as above.
ln x = (x-1) - (x-1)^2/2 + (x-1)^3/3 + …
Taylor about a=π/4: f(a)=sin(π/4)=√2/2, f'(a)=cos(π/4)=√2/2, f''(a)=-sin(π/4)=-√2/2, f'''(a)=-cos(π/4)=-√2/2. Hence sin x = √2/2 + √2/2·h + (-√2/2)·h^2/2 + (-√2/2)·h^3/6 + … where h=x-π/4. Simplify coefficients to obtain terms shown.
sin x = \frac{\sqrt2}{2} + \frac{\sqrt2}{2}(x-\frac{\pi}{4}) - \frac{\sqrt2}{4}(x-\frac{\pi}{4})^2 + … (first three non-zero terms include also the cubic term -\frac{\sqrt2}{12}(x-\frac{\pi}{4})^3 ).
Let h=x-1 ⇒ x=h+1. Then f=(h+1)^2 -3(h+1)+2 = h^2 - h. Hence f(x)=(x-1)^2 - (x-1).
f(x)=(x-1)^2 - (x-1).
(1) Use Taylor: cos x =1 - x^2/2 + … so (cos x -1)/x^2 → -1/2. (2) Leading terms: ratio → coefficient ratio 1/3. (3) ln x grows slower than x so (ln x)/x→0 (or use L'Hôpital: derivative 1/x over 1 gives 0).
(1) -1/2. (2) 1/3. (3) 0.
With nominal annual rate r and continuous compounding, A = lim_{n→∞} A_0(1 + r/n)^{nt} = A_0 e^{rt} since lim_{m→∞} (1+u/m)^m = e^u with m= n/t and u=rt.
A(t)=A_0 e^{rt}.
Use identity sec x - tan x = (1 - sin x)/cos x = cos x/(1+sin x). As x→π/2, cos x→0 and denominator →2, so the limit is 0.
0.
(i) Exponential dominates polynomial: x/e^x→0 (L'Hôpital if desired). (ii) Standard limit: sin x ~ x as x→0, so ratio→1.
(i) 0. (ii) 1.
(7) Let h=x-1 →0^+, then x^{1/(x-1)}=(1+h)^{1/h}→e. (8) Standard limit (1+1/x)^x→e as x→∞.
(7) e. (8) e.
Applying L’ Hôpital’s Rule
Exponentiating we get, $\lim _{x \rightarrow ∞}$ g(x) = e 1 = e
Applying L’ Hôpital’s Rule Exponentiating we get, $\lim _{x \rightarrow ∞}$ g(x) = e 1 = e
$\lim _{x \rightarrow \frac{π}{2}}$ (sin x) tan x [1 ∞ indeterminate form]
Let g(x) = (sin x) tan x
Taking log on both sides,
log g(x) = tan x log sin x
$\lim _{x \rightarrow \frac{π}{2}}$ log g(x) = $\lim _{x \rightarrow \frac{π}{2}}$ $\frac { log sin x }{ cot x }$
[ $\frac { 0 }{ 0 }$ Indeterminate form
Applying L’ Hôpital’s Rule
= $\lim _{x \rightarrow \frac{π}{2}}$ ($\frac { cotx }{ -cosec^2x }$) = -1
exponentiating, we get
$\lim _{x \rightarrow \frac{π}{2}}$ g(x) = e -1 = $\frac { 1 }{ e }$
= $\lim _{x \rightarrow \frac{π}{2}}$ ($\frac { cotx }{ -cosec^2x }$) = -1 exponentiating, we get $\lim _{x \rightarrow \frac{π}{2}}$ g(x) = e -1 = $\frac { 1 }{ e }$
$\lim _{x \rightarrow 0^+}$ (cos x) $\frac { 1 }{ x^2 }$ [1 ∞ indeterminate form
let g(x) = (cos x) $\frac { 1 }{ x^2 }$
Taking log on both sides,
$\lim _{x \rightarrow 0^+}$ (cos x) $\frac { 1 }{ x^2 }$ [1 ∞ indeterminate form let g(x) = (cos x) $\frac { 1 }{ x^2 }$ Taking log on both sides,
Set \(A=A_0\left(1+\dfrac{r}{n}\right)^{nt}=A_0\left[\left(1+\dfrac{r}{n}\right)^{n/r}\right]^{rt}\). As \(n\to\infty\), \(\left(1+\dfrac{r}{n}\right)^{n/r}\to e\). Hence limit is \(A_0 e^{rt}\).
A_0 e^{rt}
(i) f(x) = x² – 12x + 10; [1, 2]
(ii) f(x) = 3x 4 – 4x³; [-1, 2].
(iii) f(x)= 6x $\frac { 4 }{ 3 }$ – 3x $\frac { 1 }{ 3 }$; [-1, 1]
(iv) f(x) = 2 cos x + sin 2x; [0, $\frac { π }{ 2 }$ ]v
(i) f(x) = x² – 12x + 10;
f'(x) = 2x – 12
f'(x) = 0 ⇒ 2x – 12 = 0
x = 6 ∉ (1, 2)
Now, Evaluating f(x) at the end points x = 1, 2
f(1) = 1 – 12 + 10 = -1
f(2) = 4 – 24 + 10 = -10
Absolute maximum f(1) = -1
Absolute minimum f(2) = -10
(ii) f(x) = 3x 4 – 4x 3
f'(x) = 12x 3 – 12x 2
f'(x) = 0 ⇒ 12x 2 (x – 1) = 0
⇒ x = 0 or x = 1
[Here x = 0, 1 ∈ [-1, 2]]
Now f (-1) = 4
f(0) = 0
f(1) = -1
f(2) = 16
so absolute maximum = 16 and absolute minimum = -1
f(1) = -1 f(2) = 16 so absolute maximum = 16 and absolute minimum = -1
(i) f(x) = 2x³ + 3x² – 12x
(ii) f(x) = $\frac { x }{ x-5 }$
(iii) f(x) = $\frac { e^x }{ 1-e^x }$
(iv) f(x) = $\frac { x^3 }{ 3 }$ – log x
(v) f(x) = sin x cos x+ 5, x ∈ (0, 2π)v
(i) f(x) = 2x³ + 3x² – 12x
f'(x) = 6x² + 6x – 12
f'(x) = 0 ⇒ 6(x² + x – 2) = 0
(x + 2)(x – 1) = 0
Stationary points x = -2, 1
Now, the intervals of monotonicity are
(-∞, -2), (-2, 1) and (1, ∞)
In (-∞, -2), f'(x) > 0 ⇒ f(x) is strictly increasing.
In (-2, 1), f'(x) < 0 ⇒ f(x) is strictly decreasing.
In (1, ∞), f'(x) > 0 ⇒ f(x) is strictly increasing.
f(x) attains local maximum as f'(x) changes its sign from positive to negative when passing through x = -2.
Local maximum
f(-2) = 2 (-8) + 3 (4) – 12 (-2)
= -16 + 12 + 24 = 20
f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = 1.
∴ Local minimum f(1) = 2 + 3 – 12 = -7
(ii) f(x) = $\frac { x }{ x-5 }$
f'(x) = $\frac { (x-5)(1)-x(1) }{ (x-5)^2 }$ = –$\frac { 5 }{ (x-5)^2 }$
f'(x) = 0, which is absured
But in f(x) = $\frac { x }{ x-5 }$
The function is defined only when x < 5 or x > 5
∴ The intervals are (-∞, 5) and (5, ∞)
In the interval (-∞, 5), f'(x) < 0
In the interval (5, ∞), f'(x) < 0
∴ f(x) is strictly decreasing in (-∞, 5) and (5, ∞)
When x = 0, f(x) becomes undefined.
∴ x = 0 is an excluded value.
∴ The intervals are (-∞, 0) ∪ (0, ∞) in – (-∞, ∞), f'(x) > 0
∴ f(x) is strictly increasing in (- ∞, ∞) and there is no extremum.
(iv) f(x)= $\frac { x^3 }{ 3 }$ – log x
f'(x) = x² – $\frac { 1 }{ x }$
f'(x) = 0 ⇒ x³ – 1 = 0 ⇒ x = 1
The intervals are (0, 1) and (1, ∞).
i.e., when x > 0, the function f(x) is defined in the interval (0, 1), f'(x) < 0
∴ f(x) is strictly decreasing in (0, 1) in the interval (1, ∞), f'(x) > 0
∴f(x) is strictly increasing in (1, ∞)
f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = 1
∴ Local minimum
f(1) = $\frac { 1 }{ 3 }$ – log 1 = $\frac { 1 }{ 3 }$ – 0 = $\frac { 1 }{ 3 }$
(v) f(x) = sin x cos x + 5, x ∈ (0, 2π)
f'(x) = cos 2x
f'(x) = 0 ⇒ cos 2x = 0
Stationary points
x = $\frac { π }{ 4 }$, $\frac { 3π }{ 4 }$, $\frac { 5π }{ 4 }$, $\frac { π }{ 4 }$ ∈x = (0, 2π)
In the interval (0, $\frac { π }{ 4 }$), f'(x) > 0 ⇒ f(x) is strictly increasing.
In the interval ($\frac { π }{ 4 }$, $\frac { 3π }{ 4 }$), f'(x) < 0 ⇒ f(x) is strictly decreasing.
In the interval ($\frac { 3π }{ 4 }$, $\frac { 5π }{ 4 }$), f'(x) > 0 ⇒ f(x) is strictly increasing.
In the interval ($\frac { 5π }{ 4 }$, $\frac { 7π }{ 4 }$), f'(x) < 0 ⇒ f(x) is strictly decreasing.
In the interval ($\frac { 7π }{ 4 }$, 2π), f'(x) > 0 ⇒ f(x) is strictly increasing.
f'(x) changes its sign from positive to negative when passing through x = $\frac { π }{ 4 }$ and x = $\frac { 5π }{ 4 }$
∴ f(x) attains local maximum at x = $\frac { π }{ 4 }$ and $\frac { 5π }{ 4 }$
f'(x) changes its sign from negative to positive when passing through x = $\frac { 3π }{ 4 }$ and x = $\frac { 7π }{ 4 }$
∴ f(x) attains local maximum at x = $\frac { 3π }{ 4 }$ and x = $\frac { 5π }{ 4 }$
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KOTAKBANK Pivot Point Calculator
∴ f(x) attains local maximum at x = $\frac { 3π }{ 4 }$ and x = $\frac { 5π }{ 4 }$ Read More: KOTAKBANK Pivot Point Calculator
(i) f(x) = x(x – 4)³
(ii) f(x) = sin x + cos x, 0 < x < 2π
(iii) f(x) = $\frac { 1 }{ 2 }$(e x – e -x )v
(i) f(x) = x(x – 4)³
f'(x) = 3x(x – 4)² + (x – 4)³(1)
= (x – 4)² (4x – 4) = 4 (x – 4)² (x – 1)
f'(x) = 4 [(x-4)² (1) + (x – 1) 2 (x – 4)]
= 4(x – 4)(x – 4 + 2x – 2)
= 4(x – 4)(3x – 6) = 12(x – 4)(x – 2)
f'(x) = 0 ⇒ 12 (x – 4) (x – 2) = 0
Critical points x = 2, 4
The intervals are (- ∞, 2), (2, 4) and (4, ∞)
In the interval (-∞, 2), f”(x) > 0 ⇒ Curve is Concave upward
In the interval (2, 4), f”(x) < 0 ⇒ Curve is Concave downward.
In the interval (4, ∞), f”(x) > 0 ⇒ Curve is Concave upward.
The curve is concave upward in
(-∞, 2), (4, ∞) it is concave downward in (2, 4).
f”(x) changes its sign when passing through x = 2 and x = 4
Now f(2) = 2 (2 – 4)³ = -16 and f(4) = 4 (4 – 4)³ = 0
∴ The points of inflection are (2, -16) and (4, 0).
(ii) f(x) = sin x + cos x, 0 < x < 2π
f'(x) = cos x – sin x
f”(x) = – sin x – cos x
f'(x) = 0 ⇒ sin x + cos x = 0
Critical points x = $\frac { 3π }{ 4 }$, $\frac { 7π }{ 4 }$
The intervals are (0, $\frac { 3π }{ 4 }$), ($\frac { 3π }{ 4 }$, $\frac { 7π }{ 4 }$) and ($\frac { 7π }{ 4 }$, 2π)
In the interval (0, $\frac { 3π }{ 4 }$), f'(x) < 0 ⇒ curve is concave down.
In the interval ($\frac { 3π }{ 4 }$, $\frac { 7π }{ 4 }$), f'(x) > 0 ⇒ curve is concave up.
In the interval ($\frac { 7π }{ 4 }$, 2π), f'(x) < 0 ⇒ curve is concave down.
The curve is concave upward in ($\frac { 3π }{ 4 }$, $\frac { 7π }{ 4 }$) and concave downward in (0, $\frac { 3π }{ 4 }$) and ($\frac { 7π }{ 4 }$, 2π)
f'(x) changes its sign when passing through x = $\frac { 3π }{ 4 }$ and x = $\frac { 7π }{ 4 }$
(iii) f(x) = $\frac { 1 }{ 2 }$ (e x – e -x )
f'(x) = $\frac { 1 }{ 2 }$ (e x + e -x )
f”(x) = $\frac { 1 }{ 2 }$ (e x – e -x )
f”(x) = 0 ⇒ $\frac { 1 }{ 2 }$ (e x – e -x ) = 0
Critical point x = 0
The intervals are (-∞, o) and (0, ∞)
In the interval (-∞, 0), f”(x) < 0 ⇒ curve is concave down.
In the interval (0, ∞), f(x) > 0 ⇒ curve is concave up.
∴ The curve is concave up in (0, ∞) and concave down in (-∞, 0).
f'(x) changes its sign when passing through x = 0
Now f(0) = – (e° – e°) = $\frac { 1 }{ 2 }$ (1 – 1) = 0
∴ The point of inflection is (0, 0).
f'(x) changes its sign when passing through x = 0 Now f(0) = – (e° – e°) = $\frac { 1 }{ 2 }$ (1 – 1) = 0 ∴ The point of inflection is (0, 0).
(i) f(x) = -3x 5 + 5x 3
(ii) f(x) = x log x
(iii) f(x) = x² e -2xv
(i) f(x) = – 3x 5 + 5x 3
f'(x) = 0, f”(x) = -ve at x = a
⇒ x = a is a maximum point
f'(x) = 0, f”(x) = +ve at x = 6
⇒ x = b is a minimum point
f(x) = – 3x 5 + 5x 3
f’ (x) = -15x 4 + 15x 2
f”(x) = -60x 3 + 30x
f'(x) = 0 ⇒ – 15x 2 (x 2 – 1) = 0
⇒ x = 0, +1, -1
at x = 0, f”(x) = 0
at x = 1, f”(x) = -60 + 30 = – ve
at x = -1, f”(x) = 60 – 30 = + ve
So at x = 1, f'(x) = 0 and f”(x) = -ve
⇒ x = 1 is a local maximum point.
and f(1) = 2
So the local maximum is (1, 2)
at x = -1, f'(x) = 0 and f”(x) = +ve
⇒ x = -1 is a local maximum point and f(-1) = -2.
So the local minimum point is (-1, -2)
∴ a local minimum is -2 and the local maximum is 2.
(ii) f(x) = x log x
f'(x) = x – $\frac { 1 }{ 4 }$ + log x = 1 + log x
For maximum or minimum
f'(x) = 0 ⇒ 1 + log x = 0
⇒ log x = -1
x = e -1 = $\frac { 1 }{ e }$
f”(x) = $\frac { 1 }{ x }$
at x = $\frac { 1 }{ e }$, f”(x) > 0 ⇒ f(x) attains minimum.
∴ Local minimum f($\frac { 1 }{ e }$) = $\frac { 1 }{ e }$ log ($\frac { 1 }{ e }$)
= $\frac { 1 }{ e }$(-1) = –$\frac { 1 }{ e }$
(iii) f(x) = x 2 e -2x
f'(x) = x 2 [-2e -2x ] + e -2x (2x)
= 2e -2x (x – x 2 )
f”(x) = 2e -2x (1 – 2x) + (x – 2 ) (-4e -2x )
= 2e -2x [(1 – 2x) + (x – x 2 ) (- 2)]
= 2e -2x [2x 2 – 4x + 1]
f'(x) = 0 ⇒ 2e -2x (x – x 2 ) = 0
⇒ x (1 – x) = 0
⇒ x = 0 or x = 1
at x = 0, f”(x) = 2 × 1 [0 – 0 + 1] = +ve
⇒ x = 0 is a local minimum point and the minimum value is f(0) = 0 at x = 1,
f”(x) = 2e -2 [2 – 4 + 1] = -ve
⇒ x = 1 is a local maximum point and the maximum value is f(1) = $\frac{1}{e^{2}}$
Local maxima $\frac{1}{e^{2}}$ and local minima = 0
f”(x) = 2e -2 [2 – 4 + 1] = -ve ⇒ x = 1 is a local maximum point and the maximum value is f(1) = $\frac{1}{e^{2}}$ Local maxima $\frac{1}{e^{2}}$ and local minima = 0
(x) = 4x³ + 3x² – 6x + 1
Monotonicity
f(x) = 4x³ + 3x² – 6x + 1
f'(x) = 12x² + 6x – 6
f'(x) = 0 ⇒ 6(2x² + x – 1) = 0
x = -1, $\frac { 1 }{ 2 }$ (Stationary points)
∴ The intervals of monotonicity are (-∞, -1), (-1, $\frac { 1 }{ 2 }$) and ($\frac { 1 }{ 2 }$, ∞)
In (-∞, -1), f'(x) > 0 ⇒ f(x) is strictly increasing
In (-1, $\frac { 1 }{ 2 }$), f'(x) < 0 ⇒ f(x) is strictly decreasing
In ($\frac { 1 }{ 2 }$, ∞) f'(x) > 0 ⇒ f(x) is strictly increasing
f(x) attains local maximum as f'(x) changes its sign from positive to negative when passing through x = -1
∴ Local maximum f(-1) = -4 + 3 + 6 + 1 = 6
f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = $\frac { 1 }{ 2 }$
∴ Local minimum f($\frac { 1 }{ 2 }$)
= 4($\frac { 1 }{ 8 }$) + 3($\frac { 1 }{ 4 }$) – 6($\frac { 1 }{ 2 }$) + 1
= $\frac { 1 }{ 2 }$ + $\frac { 3 }{ 4 }$ – + 1 = –$\frac { 3 }{ 4 }$
f(x) = 4x³ + 3x² – 6x + 1
f'(x) = 12x² + 6x – 6
f”(x) = 24x + 6
f’(x) = 0 ⇒ 24x + 6 = 0
x = –$\frac { 6 }{ 24 }$ = –$\frac { 1 }{ 4 }$ (critical points)
∴ The intervals are (∞, $\frac { 1 }{ 4 }$) and ($\frac { 1 }{ 4 }$, ∞) f”(x) > 0
In the interval (-∞, –$\frac { 1 }{ 4 }$), f”(x) < 0 ⇒ curve is concave down.
In the interval (-$\frac { 1 }{ 4 }$, ∞), f”(x) > 0 ⇒ curve is concave up.
The curve is concave upward in (-$\frac { 1 }{ 4 }$, ∞) and concave downward in (-∞, –$\frac { 1 }{ 4 }$)
f”(x) changes its sign when passing through x = –$\frac { 1 }{ 4 }$
EICHERMOT Pivot Calculator
The curve is concave upward in (-$\frac { 1 }{ 4 }$, ∞) and concave downward in (-∞, –$\frac { 1 }{ 4 }$) f”(x) changes its sign when passing through x = –$\frac { 1 }{ 4 }$ EICHERMOT Pivot Calculator
Let numbers be \(x\) and \(12-x\). Product \(P=x(12-x)=12x-x^2\). \(P'=12-2x=0\Rightarrow x=6\). So numbers are \(6,6\); maximum product \(36\).
6 and 6
With product fixed, sum is minimized when numbers are equal: \(x=y=\sqrt{20}=2\sqrt5\). Sum = \(4\sqrt5\).
Both numbers = 2\sqrt{5}; minimum sum = 4\sqrt{5}
Let \(h=6-r\). Then \(V=\pi r^2(6-r)=\pi(6r^2-r^3)\). \(V'=\pi(12r-3r^2)=3\pi r(4-r)=0\Rightarrow r=0,4\). \(r=4\) gives max \(V=\pi\cdot16\cdot2=32\pi\). At boundary \(r=0\) (or \(r=6\)) volume is 0 (min).
Maximum \(V=32\pi\) at \(r=4,h=2\); minimum \(V=0\) at boundary \(r=0\) (or \(h=0\)).
Minimize \(x^2+y^2=(x+y)^2-2xy=100-2xy\). To minimize, maximize \(xy\) which occurs at \(x=y=5\). Then \(x^2+y^2=2\cdot25=50\).
50
Perimeter 2(l+w)=40 so l+w=20. Area \(A=lw\) is maximized when \(l=w=10\) (AM-GM). Max area =100 m^2.
100 m^2 (square 10 m × 10 m)
Let printed width \(x\) and printed height \(y\) with \(xy=24\). Page width \(W=x+2\), height \(H=y+3\). Area \(A=(x+2)(y+3)=30+3x+48/x\). \(A'=3-48/x^2=0\Rightarrow x^2=16\Rightarrow x=4\), so \(y=6\). Thus \(W=6\) cm, \(H=9\) cm.
Page width = 6 cm, height = 9 cm
Let width along river = x, depth = y, with xy=180000. Fencing required \(P=x+2y=x+360000/x\). \(P'=1-360000/x^2=0\Rightarrow x=600\). Then \(y=300\) and fencing \(600+2\cdot300=1200\) m.
Minimum fencing = 1200 m, with dimensions 600 m (along river) by 300 m (depth).
For rectangle with half-sides \(x,y\), constraint \(x^2+y^2=100\). Area \(A=4xy\le 4\cdot\dfrac{x^2+y^2}{2}=4\cdot50=200\) (AM-GM). Equality when \(x=y=5\sqrt2\); full side = \(2x=10\sqrt2\).
Square of side \(10\sqrt2\) cm; max area = 200 cm^2
Let sides be \(x,y\) with perimeter fixed: \(x+y=\dfrac{P}{2}=c\). Area \(A=xy\). Using AM-GM, \(\dfrac{x+y}{2}\ge\sqrt{xy}\Rightarrow xy\le\left(\dfrac{c}{2}\right)^2\). Equality when \(x=y\). Hence the square gives maximum area.
Square maximizes area.
Let half-width \(x\) and height \(y\) with \(x^2+y^2=r^2\). Area \(A=2xy\). Maximize: \(A=2x\sqrt{r^2-x^2}\). Set derivative zero ⇒ \(r^2-2x^2=0\) ⇒ \(x=r/\sqrt2\), \(y=r/\sqrt2\). So full width \(2x=\sqrt2 r\), height \(r/\sqrt2\), area \(=r^2\).
Width = \(\sqrt2\,r\), height = \(r/\sqrt2\) (so half-width \(r/\sqrt2\)). Max area = \(r^2\).
Surface area (open) \(S=a^2+4ah=108\). So \(h=(108-a^2)/(4a)\). Volume \(V=a^2h=\dfrac{a(108-a^2)}{4}\). \(V'=\dfrac{108-3a^2}{4}=0\Rightarrow a^2=36\Rightarrow a=6\). Then \(h=(108-36)/(24)=3\). Volume \(=36\cdot3=108\) cm^3.
Base side \(a=6\) cm, height \(h=3\) cm (max volume 108 cm^3).
Total surface area S = 2\pi r^2 + 2\pi r h. With fixed V, h = V/(\pi r^2). So S(r)=2\pi r^2 + 2\pi r \cdot \frac{V}{\pi r^2}=2\pi r^2 + \frac{2V}{r}. Differentiate: S'(r)=4\pi r - \frac{2V}{r^2}. Set S'(r)=0 => 4\pi r = 2V/r^2 => 2\pi r^3 = V. But V=\pi r^2 h, so \pi r^2 h = 2\pi r^3 => h = 2r. S''(r)=4\pi + \frac{4V}{r^3}>0 so minimum attained.
h = 2r
Let the inscribed cylinder have radius r and height h. By similarity of triangles, r/a = (b - h)/b => r = a(1 - h/b). Cylinder volume V(h)=\pi r^2 h = \pi a^2(1 - h/b)^2 h. Put u=h/b, 0<u<1. Then V(u)=\pi a^2 b\,u(1-u)^2. Maximise f(u)=u(1-u)^2: f'(u)=(1-u)(1-3u). Roots u=1 (min) and u=1/3 (max). So h=b/3 and r=a(1-1/3)=2a/3. Hence V_cyl=\pi (2a/3)^2(b/3)=\pi a^2 b (4/27). Cone volume V_cone=(1/3)\pi a^2 b = \pi a^2 b /3. Ratio = (4/27)/(1/3)=4/9.
V_cyl = \tfrac{4}{9} V_{cone}
(i) f(x) = $\frac { x^2 }{ x^2-1 }$
(ii) f(x) = $\frac { x^2 }{ x+1 }$
(iii) f(x) = $\frac { 3x }{ \sqrt{x^2+2} }$
(iv) f(x) = $\frac { x^2-6x-1 }{ x+3 }$
(v) f(x) = $\frac { x^2+6x-4 }{ 3x-6 }$v
(i) f(x) = $\frac { x^2 }{ x^2-1 }$
The function becomes undefined when x = 1 and x = -1
∴ x = 1 and x = -1 are the vertical asymptotes
As ‘x’ gets larger (Positive or negative) the function, the function attaining the value 1.
∴ y = 1 is horizontal asymptote
(ii) f(x) = $\frac { x^2 }{ x+1 }$
The function becomes undefined when x = -1
∴ Vertical asymptote is x = -1 and there is no horizontal asymptote.
No horizontal asymptote exists for the curve. Oblique asymptote can be obtained by polynomial long division method.
Oblique (or) slant asymptote is y = x – 1
(iii) f(x) = $\frac { 3x }{ \sqrt{x^2+2} }$
No vertical asymptotes
Horizontal asymptotes RHL (Right Hand Limit)
y = 3 and y = -3 are the Horizontal asymptotes
Slant asymptotes’. No such slant asymptotes exist for the given curve.
(iv) f(x) = $\frac { x^2-6x-1 }{ x+3 }$
When x = -3, the function becomes undefined.
∴ x = -3 is the vertical asymptote.
No Horizontal asymptote exist for the curve.
Oblique asymptote can be obtained by polynomial long division method
∴ y = x – 9 is the slant (or) oblique asymptote.
(v) f(x) = $\frac { x^2+6x-4 }{ 3x-6 }$
The function becomes undefined when x = 2.
∴ x = 2 is the vertical asymptote.
No Horizontal asymptote exist for the given curve.
Oblique asymptote can be obtained by polynomial long division method.
∴ y = $\frac { x }{ 3 }$ + $\frac { 8 }{ 3 }$ (or) 3y = x + 8 is the slant asymptote.
No Horizontal asymptote exist for the given curve. Oblique asymptote can be obtained by polynomial long division method. ∴ y = $\frac { x }{ 3 }$ + $\frac { 8 }{ 3 }$ (or) 3y = x + 8 is the slant asymptote.
(i) y = –$\frac { 1 }{ 3 }$ (x³ – 3x + 2)
(ii) y = x $\sqrt { 4-x }$
(iii) y = $\frac { x^2+1 }{ x^2-4 }$
(iv) y = $\frac { 1 }{ 1+e^{-x} }$
(v) y = $\frac { x^3 }{ 24 }$ – log xv
(i) y = –$\frac { 1 }{ 3 }$ (x³ – 3x + 2)
Factorizing we get
y = –$\frac { 1 }{ 3 }$ (x – 1)² (x + 2) = f(x)
The domain and the range of the given function f(x) are the entire real line.
Putting y = 0, we get x = 1, 1, – 2. Hence the x-intercepts are (1, 0) and (- 2, 0) and by putting x = 0. We get y = –$\frac { 2 }{ 3 }$. Therefore, the y-intercept is (0, –$\frac { 2 }{ 3 }$)
f'(x) = $\frac { (3x^2-3) }{ 3 }$ = -(x² – 1) = 1 – x²
f'(x) = 0 ⇒ 1 – x² = 0 ⇒ x = ±1
The critical points of the curve occur at x = ± 1.
f”(x) = -2x
f”(1) = – 2 < 0, ∴ f(x) is maximum at x = 1 and the local maximum is f(1) = o
f”(-1) = 2 > 0, ∴ f(x) is minimum at x = -1 and the local minimum is
f(-1) = –$\frac { 4 }{ 3 }$
f”(x) = – 2x < 0 ∀ x > 0, ∴ The function is concave downward in the positive real line.
f”(x) = 2x > 0 ∀ x < 0, ∴ The function is concave upward in the negative real line.
Since f”(x) = 0 at x = 0 and f”(x) changes its sign when passing through x = 0.
Hence the point of inflection is (0, –$\frac { 2 }{ 3 }$)
The curve has no asymptotes.
(ii) y = x$\sqrt { 4-x }$
y = x$\sqrt { 4-x }$ = f(x)
where x > 4 the curve does not exist and it exists for x ≤ 4
∴ The domain is (-∞, 4] and the Range is (-∞, $\frac { 16 }{ 3√3 }$ ]
The curve passes through the origin. The curve intersects x-axis at (4, 0).
f'(x) = –$\frac { x }{ 2 \sqrt{4-x} }$ + $\sqrt { 4-x }$ = $\frac { 8-3x }{ 2 \sqrt{4-x} }$
f'(x) = 0 ⇒ 8 – 3x = 0 ⇒ x = $\frac { 8 }{ 3 }$
∴ Critical point of the curve occur at x = $\frac { 8 }{ 3 }$
f”(x) = $\frac { 3x-16 }{ 4(4-x)^{\frac{3}{2}} }$
f”($\frac { 8 }{ 3 }$) = –$\frac { 3√3 }{ 4 }$ < 0
∴ f(x) is maximum at x = $\frac { 8 }{ 3 }$ and the local maximum f($\frac { 8 }{ 3 }$) = $\frac { 16 }{ 3√3 }$ and local minimum is 0 at x = 4 (from the graph)
f”(x) = $\frac { 3x-16 }{ 4(4-x)^{\frac{3}{2}} }$ < 0 ∀ x < 4
∴ The curve is concave downward in the negative real line.
No point of inflection exists.
As x → ∞, y → ±∞, and hence the curve does not have any asymptotes.
(iii) y = $\frac { x^2+1 }{ x^2-4 }$
The domain of the given function f(x) is (-∞, -2) ∪ (-2, 2) ∪ (2, ∞)
ie. x < -2 (or) -2 < x < 2 (or) x > 2.
Range of f(x) is (-∞, –$\frac { 1 }{ 4 }$) ∪ (1, ∞)
i.,e. f(x) ≤ –$\frac { 1 }{ 4 }$ (or) f(x) > 1.
Putting y = 0, x is unreal. Hence, there is no ‘x’ intercept.
By putting x = 0, we get y = –$\frac { 1 }{ 4 }$.
∴ y intercept is (0, –$\frac { 1 }{ 4 }$)
f'(x) = –$\frac { 10x }{ (x^2-4)^2 }$
f'(x) = 0 ⇒ x = 0,
∴ The critical point is at x = 0
f'(x) = $\frac { 10(x^2-4)(3x^+4) }{ (x^2-4)^4 }$
f'(0) = –$\frac { 5 }{ 8 }$ < 0,
∴ f(x) is maximum at
x = 0. Hence the local maximum is f(0) = –$\frac { 1 }{ 4 }$
No points of inflection exist for the curve.
When x = ± 2, y = ∞
∴ Vertical asymptotes are x = 2 and x = -2 and Horizontal asymptote is y = 1.
(iv) y = $\frac { 1 }{ 1+e^{-x} }$
The Domain of the function f(x) is the entire real line.
ie., (-∞, ∞) ⇒ -∞ < x < ∞ and the range is (0, 1) ie., 0 < f(x) < 1
No ‘x’ intercept for f(x) and when x = 0
y = $\frac { 1 }{ 2 }$
∴ The ‘y’ intercept is (0, $\frac { 1 }{ 2 }$)
f'(x) = $\frac { e^{-x} }{ (1+e^{-x})^2 }$
f'(x) = 0 ⇒ which is absurd. Hence there is no extremum.
No vertical asymptote for the curve exist and the Horizontal asymptotes are y = 1 and y = 0.
(v) y = $\frac { x^3 }{ 24 }$ – log x
The curve exists only for positive values of ‘x’ (x > 0) ie., domain is (0, ∞) and
The range is ($\frac { 1 }{ 3 }$ – log e², ∞)
No intersection points are possible
f'(x) = $\frac { x^2 }{ 8 }$ – $\frac { 1 }{ x }$
f'(x) = 0 ⇒ x³ – 8 = 0 ⇒ x = 2
∴ Critical point occur at x = 2
f'(x) = $\frac { x }{ 4 }$ + $\frac { 1 }{ x^2 }$
f”(2) = $\frac { 3 }{ 4 }$ > 0,
∴ f(x) is mimmum at x = 2 and the local minimum is f(2) = $\frac { 1 }{ 3 }$ – log e²
No point of inflection exists.
No Horizontal asymptotes are possible, but the vertical asymptote is x = 0 (y-axis).
ESCORTS Pivot Calculator
No point of inflection exists. No Horizontal asymptotes are possible, but the vertical asymptote is x = 0 (y-axis). ESCORTS Pivot Calculator
V=(4/3)\pi r^3, so dV/dt=4\pi r^2 dr/dt. Given dV/dt=3\pi and r=1/2: dr/dt=(3\pi)/(4\pi(1/2)^2)=3\pi/(4\pi/4)=3\pi/\pi=3 cm/s.
If y(t) is height and θ=arctan(y/40), then dθ/dt = (1/(1+(y/40)^2))*(1/40) dy/dt. With dy/dt=10 and y=30: dθ/dt = (1600/(1600+900))*(10/40)=400/2500=4/25 rad/s.
(A) t = 0 (B) t = $\frac { 1 }{ 3 }$ (C) t = 1 (D) t = 3v
(b) t = $\frac { 1 }{ 3 }$
Hint:
s(t) = 3t² – 2t – 8
Velocity V = $\frac { ds }{ dt }$ = 6t – 2
When the particle comes to rest,
velocity V = 0
6t – 2 = 0
t = $\frac { 1 }{ 3 }$
t = $\frac { 1 }{ 3 }$
(A) t = 0 (B) t = $\frac { 1 }{ 3 }$ (C) t = 1 (D) t = 3v
(b) t = $\frac { 1 }{ 3 }$
Hint:
s(t) = 3t² – 2t – 8
Velocity V = $\frac { ds }{ dt }$ = 6t – 2
When the particle comes to rest,
velocity V = 0
6t – 2 = 0
t = $\frac { 1 }{ 3 }$
t = $\frac { 1 }{ 3 }$
Velocity v=dx/dt=80 -32t. Set v=0 => 80 -32t=0 => t=80/32=2.5 s.
(A) (4, 11) (B) (4, -11) (C) (-4, 11) (D) (-4, -11)v
(a) (4, 11)
Hint:
x = ±4
When x = 4, y = 11
∴ Point on the curve is (4, 11).
(4, 11)
(A) -8 (B) -4 (C) -2 (D) 0v
(b) -4
Hint:
f(x) = $\sqrt { 8-2x }$
f'(x) = –$\frac { 2 }{ 2\sqrt { 8-2x } }$ = –$\frac { 1 }{ \sqrt { 8-2x } }$
Slope of the tangent is – 0.25
ie„ f'(x) = -0.25
–$\frac { 1 }{ \sqrt { 8-2x } }$ = -0.25 = $\frac { -1 }{ 4 }$
$\sqrt { 8-2x }$ = 4
8 – 2x = 16
x = –$\frac { 8 }{ 2 }$ = -4
Abscissa x = -4
-4
(A) -4√3 (B) -4 (C) –$\frac { √3 }{ 12 }$ (D) 4√3v
(c) –$\frac { √3 }{ 12 }$
Hint:
f(x) = 2 cos 4x
f'(x) = -8 sin 4x
Slope of the normal at x = $\frac { π }{ 12 }$ is
–$\frac { √3 }{ 12 }$
(A) y = 0 (B) y = ±√3 (C) –$\frac { 1 }{ 2 }$ (D) y = ±3v
(d) y = ±3
Hint:
y² – xy + 9 = 0 ……… (1)
2y$\frac { dy }{ dx }$ – (x$\frac { dy }{ dx }$ + y) = 0
$\frac { dy }{ dx }$ (2y – x) = y
$\frac { dy }{ dx }$ = $\frac { y }{ 2y-x }$
When the tangent is vertical $\frac { dy }{ dx }$ = ∞
i.e., $\frac { y }{ 2y-x }$ = $\frac { 0 }{ 1 }$
⇒ 2y – x = 0
2y = x
sub in (1)
y² – 2y² + 9 = 0
⇒ y² = 9
y = ±3
y = ±3
At origin y=x^2 has tangent y=0 (x-axis). Curve x=y^2 has tangent x=0 (y-axis). Angle between x- and y-axis is \pi/2.
(A) 0 (B) 1 (C) 2 (D) ∞v
(d) ∞
Hint:
∞
(A) [ $\frac { 5π }{ 8 }$, $\frac { 3π }{ 4 }$ ] (B) [ $\frac { π }{ 2 }$, $\frac { 5π }{ 8 }$ ] (C) [ $\frac { π }{ 4 }$, $\frac { π }{ 2 }$ ] (D) [ 0, $\frac { π }{ 4 }$ ]v
(c) [ $\frac { π }{ 4 }$, $\frac { π }{ 2 }$ ]
Hint:
f(x) = sin 4 x + cos 4 x
f'(x) = 4 sin³ x cos x – 4 cos³ x sin x
f'(x) = 0 ⇒ 4 sin x cos x (sin²x – cos²x) = 0
sin x = 0; cos x = 0; sin² – cos² x = 0
x = 0; x = $\frac { π }{ 2 }$; sin² x = cos² x
x = $\frac { π }{ 4 }$
In [0, $\frac { π }{ 2 }$ ], f'(x) = -ve ⇒ f(x) is decreasing
In [ $\frac { π }{ 2 }$, $\frac { π }{ 2 }$ ], f'(x) = +ve ⇒ f(x) is increasing
[ $\frac { π }{ 4 }$, $\frac { π }{ 2 }$ ]
(A) 1 (B) √2 (C) $\frac { 3 }{ 2 }$ (D) 2v
(d) 2
Hint:
f(x) = x³ – 3x²
f'(x) = 3x² – 6x
f'(x) = 0
⇒ 3x (x – 2) = 0
x = 0, 2
2
(A) 2 (B) 2.5 (C) 3 (D) 3.5v
(c) 3
Hint:
3
(A) $0$ (B) $3$ (C) $6$ (D) $9$v
(d) $9$
$|3-x|\ge0$ for every $x$, attaining its minimum $0$ at $x=3$. Hence the least value of $f(x)=|3-x|+9$ is $0+9=9$ (the function has no maximum).
$9$
(A) x = $\frac { π }{ 4 }$ (B) x = $\frac { π }{ 2 }$ (C) x = π (D) x = $\frac { 3π }{ 2 }$v
(b) x = $\frac { π }{ 2 }$
Hint:
y = e x sin x, x ∈ [0, 2π] dy
Slope ‘S’ = $\frac { dy }{ dx }$ = e x cos x + e x sin x
S = e x (cos x + sin x)
$\frac { dS }{ dx }$ = e x (-sin x + cos x) + (cos x + sin x)e x
= e x (2 cos x)
For maximum or minimum,
$\frac { dS }{ dx }$ = 0 ⇒ 2e x cos x = 0
e x = 0 is not possible
∴ cos x = 0
x = $\frac { π }{ 2 }$
x = $\frac { π }{ 2 }$
Answer: The speed is 2π km/s toward the foot. Explanation: Let θ be the angle between the beam and the shore. The point where the beam hits the shore is at distance s = 5 cot θ from the perpendicular foot. ds/dt = -5 csc^2θ · dθ/dt. dθ/dt = 2π/10 = π/5 rad/s. At θ = 45° (π/4), csc^2θ = 2, so ds/dt = -5·2·(π/5) = -2π km/s. The magnitude of the speed is 2π km/s (negative sign indicates direction toward the foot).
Answer: dh/dt = 9/(10π) m/min. Explanation: Similar triangles give r = (5/12)h. Volume V = (1/3)π r^2 h = (25/432)π h^3. dV/dt = (25/144)π h^2 dh/dt. Thus dh/dt = (dV/dt)/[(25/144)π h^2] = 10 / [(25/144)π · 64] = 9/(10π) m/min.
Let numbers be x,y>0 with x^2+y^2=200. Maximize P=xy. By AM-GM or symmetry, maximum when x=y ⇒ 2x^2=200 ⇒ x=y=10 ⇒ P=100.
y'=4ax^3+2bx ⇒ horizontal tangents at x=0 or x^2=−b/(2a). Since ab>0, −b/(2a)<0 so only real root x=0 exists (so horizontal tangent exists). y''=12ax^2+2b has same sign for all x because a and b have same sign ⇒ no change of concavity ⇒ no inflection points. Hence (4).
y=x^{1/3}. y''=−(2/9)x^{−5/3}, which changes sign on crossing x=0 (undefined at 0 but sign change occurs), so the inflection point is at the origin (0,0).