What is the extreme f(x)=8x^3+81x^2-42x-8 ?

The extreme f(x)=8x^3+81x^2-42x-8 is Maximum(-7,1511),Minimum(1/4 ,-213/16)

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extreme f(x)=8x^3+81x^2-42x-8

extreme points

f (x) = 8 x^{3} + 81 x^{2} - 42 x - 8

Maximum (- 7, 1511), Minimum (\frac{1}{4}, - \frac{213}{16})

Solution steps

f^{'} (x) = 24 x^{2} + 162 x - 42

Find intervals:

Increasing

: - \infty < x < - 7,

Decreasing

: - 7 < x < \frac{1}{4},

Increasing

: \frac{1}{4} < x < \infty

Plug

x = - 7

into

8 x^{3} + 81 x^{2} - 42 x - 8 : 1511

Maximum (- 7, 1511)

Plug

x = \frac{1}{4}

into

8 x^{3} + 81 x^{2} - 42 x - 8 : - \frac{213}{16}

Minimum (\frac{1}{4}, - \frac{213}{16})

Maximum (- 7, 1511), Minimum (\frac{1}{4}, - \frac{213}{16})

x^{\frac{5}{3}} - 5 x^{\frac{2}{3}}

f (x, y) = x^{3} + 3 x y^{2} - 15 x + y^{3} - 15 y

f (x) = x + \frac{36}{x}

f (x) = \frac{x}{x ^{2} - x + 9}, 0 \leq x \leq 9

f (x) = e^{x^{2} - 3 x - 1}

What is the extreme f(x)=8x^3+81x^2-42x-8 ?
The extreme f(x)=8x^3+81x^2-42x-8 is Maximum(-7,1511),Minimum(1/4 ,-213/16)