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in Applications of Differential Calculus by (48.7k points)
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Find the local extrema for the following functions using second derivative test: 

(i) f(x) = – 3x5 + 5x3 

(ii) f(x) = x log x 

(iii) f(x) = x2 e-2x

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(i) f(x) = – 3x5 + 5x3 

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) = – 3x5 + 5x3 

f’(x) = -15x4 + 15x2 

f”(x) = -60x3 + 30x 

f'(x) = 0 ⇒ – 15x2(x2 – 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) 

∴ local minimum is -2 and local maximum is 2.

(ii) f(x) = x log x

(iii) f(x) = x2 e-2x

f'(x) = x2 [-2e-2x] + e-2x(2x) 

= 2e-2x(x – x2

f”(x) = 2e-2x(1 – 2x) + (x – x2) (-4e-2x)

= 2e-2x [(1 – 2x) + (x – x2) (- 2)] 

= 2e-2x [2x – 4x + 1] 

f'(x) = 0 ⇒ 2e-2x (x – x2) = 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) = 1/e2

Local maxima 1/eand local minima = 0

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