What is the extreme f(x)=8x^3-6x+4 ?

The extreme f(x)=8x^3-6x+4 is Maximum(-1/2 ,6),Minimum(1/2 ,2)

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extreme f(x)=8x^3-6x+4

extreme points

f (x) = 8 x^{3} - 6 x + 4

Maximum (- \frac{1}{2}, 6), Minimum (\frac{1}{2}, 2)

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 + 4 : 6

Maximum (- \frac{1}{2}, 6)

Plug

x = \frac{1}{2}

into

8 x^{3} - 6 x + 4 : 2

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

Maximum (- \frac{1}{2}, 6), Minimum (\frac{1}{2}, 2)

4 - x^{2}

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

y = \frac{x}{x ^{2} + 1}

(38.7, 38.24), (84.6, 87.04)

(- 1, 6), (- 6, - 1)

What is the extreme f(x)=8x^3-6x+4 ?
The extreme f(x)=8x^3-6x+4 is Maximum(-1/2 ,6),Minimum(1/2 ,2)