What is the extreme f(x)=8x-x^4,-2<= x<= 1 ?

The extreme f(x)=8x-x^4,-2<= x<= 1 is Minimum(-2,-32),Maximum(1,7)

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extreme f(x)=8x-x^4,-2<= x<= 1

extreme points

f (x) = 8 x - x^{4}, - 2 \leq x \leq 1

Minimum (- 2, - 32), Maximum (1, 7)

Solution steps

Find the critical points:

x = - 2, x = 1

Domain of

8 x - x^{4}, - 2 \leq x \leq 1 : - 2 \leq x \leq 1

Find intervals:

Increasing

: - 2 < x < 1

Plug the extreme point

x = - 2

into

8 x - x^{4} \Rightarrow y = - 32

Minimum (- 2, - 32)

Plug the extreme point

x = 1

into

8 x - x^{4} \Rightarrow y = 7

Maximum (1, 7)

Minimum (- 2, - 32), Maximum (1, 7)

f (x, y) = 2 x^{2} + 4 x y^{2} - 2 x y - 10 x - 2 y + 2

f (x) = 3 (x - 4) (x + 7)^{2}

f (x) = x^{3} + 2 x^{2} + x

x^{3} - 3 x^{2} + 3 x + 1

100 + \frac{1}{2} x + \frac{1800}{x}

What is the extreme f(x)=8x-x^4,-2<= x<= 1 ?
The extreme f(x)=8x-x^4,-2<= x<= 1 is Minimum(-2,-32),Maximum(1,7)