Find the coordinates of all local optimum points and stationary-inflec

1.	Find the coordinates of all local optimum points and stationary-inflection points of the following function with necessary tests:f(x) = 3x4 + 4x3 – 12x2 + 5
Answer» We have, f(x) = 3x⁴ + 4x³ - 12x² + 5 ∴ f'(x) = 12x³ + 12x² - 24x = 12x (x²+x-2) = 12x (x²+2x-x-2) = 12x (x-1)(x+2) Therefore, f'(x) = 0 gives x = 0 or x = 1 or x = -2. Hence, critical points for given function are x = 0, x = -2 and x = 1. Now, f''(x) = 36x² + 24x - 24 ∵ f''(0) = -24 < 0 ∴ x = 0 is point of local maxima (By second derivative test) Also f''(1) = 36+24-24 = 36 > 0 ∴ x = 1 is point of local minima (By second derivative test) Now, f''(-2) = 36 x 4 + 24 x -2 - 24 = 144 - 48 - 24 = 144 - 72 = 72 > 0 ∴ x = -2 is point of local minima (By second derivative test) Now, f(0) = 5, f(1) = 3+4-12+5 = 12 - 12 = 0 f(-2) = 48 - 32 - 48 + 5 = -32+5 = -27 Hence, (1, 0) and (-2, -27) are point of local minima and (0, 5) is point of local maxima.

Find the coordinates of all local optimum points and stationary-inflection points of the following function with necessary tests:f(x) = 3x4 + 4x3 – 12x2 + 5

Answer»

We have, f(x) = 3x⁴ + 4x³ - 12x² + 5

∴ f'(x) = 12x³ + 12x² - 24x

= 12x (x²+x-2)

= 12x (x²+2x-x-2)

= 12x (x-1)(x+2)

Therefore, f'(x) = 0 gives x = 0 or x = 1 or x = -2.

Hence, critical points for given function are x = 0, x = -2 and x = 1.

Now, f''(x) = 36x² + 24x - 24

∵ f''(0) = -24 < 0

∴ x = 0 is point of local maxima (By second derivative test)

Also f''(1) = 36+24-24 = 36 > 0

∴ x = 1 is point of local minima (By second derivative test)

Now, f''(-2) = 36 x 4 + 24 x -2 - 24

= 144 - 48 - 24

= 144 - 72

= 72 > 0

∴ x = -2 is point of local minima (By second derivative test)

Now, f(0) = 5, f(1) = 3+4-12+5 = 12 - 12 = 0

f(-2) = 48 - 32 - 48 + 5 = -32+5 = -27

Hence, (1, 0) and (-2, -27) are point of local minima and (0, 5) is point of local maxima.

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