What is the extreme f(x)=8x^3-6x+7 ?

The extreme f(x)=8x^3-6x+7 is Maximum(-1/2 ,9),Minimum(1/2 ,5)

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extreme f(x)=8x^3-6x+7

extreme points

f (x) = 8 x^{3} - 6 x + 7

Maximum (- \frac{1}{2}, 9), Minimum (\frac{1}{2}, 5)

Solution steps

f^{'} (x) = 24 x^{2} - 6

Find intervals:

Increasing

: - \infty < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - \frac{1}{2}

into

8 x^{3} - 6 x + 7 : 9

Maximum (- \frac{1}{2}, 9)

Plug

x = \frac{1}{2}

into

8 x^{3} - 6 x + 7 : 5

Minimum (\frac{1}{2}, 5)

Maximum (- \frac{1}{2}, 9), Minimum (\frac{1}{2}, 5)

f (x) = - (x + 1) (x - 2) (x - 3)

f (x) = 7 x^{3} - 3

(7, 6), (5, 3)

f (x) = \frac{x ^{2} + x - 6}{x ^{3} - 1}

f (x) = - x - 7

What is the extreme f(x)=8x^3-6x+7 ?
The extreme f(x)=8x^3-6x+7 is Maximum(-1/2 ,9),Minimum(1/2 ,5)