What is the extreme f(x)=-x^3+x^2-9x+8,-1<= x<= 5 ?

The extreme f(x)=-x^3+x^2-9x+8,-1<= x<= 5 is Maximum(-1,19),Minimum(5,-137)

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extreme f(x)=-x^3+x^2-9x+8,-1<= x<= 5

extreme points

f (x) = - x^{3} + x^{2} - 9 x + 8, - 1 \leq x \leq 5

Maximum (- 1, 19), Minimum (5, - 137)

Solution steps

Find the critical points:

x = - 1, x = 5

Domain of

- x^{3} + x^{2} - 9 x + 8, - 1 \leq x \leq 5 : - 1 \leq x \leq 5

Find intervals:

Decreasing

: - 1 < x < 5

Plug the extreme point

x = - 1

into

- x^{3} + x^{2} - 9 x + 8 \Rightarrow y = 19

Maximum (- 1, 19)

Plug the extreme point

x = 5

into

- x^{3} + x^{2} - 9 x + 8 \Rightarrow y = - 137

Minimum (5, - 137)

Maximum (- 1, 19), Minimum (5, - 137)

f (x) = 2 x^{3} - 6 x^{2} - 18 x + 9

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

f (x) = 2 x^{2} - 2 x^{4}

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

f (x, y) = 6 x^{2} + 2 y^{2}

What is the extreme f(x)=-x^3+x^2-9x+8,-1<= x<= 5 ?
The extreme f(x)=-x^3+x^2-9x+8,-1<= x<= 5 is Maximum(-1,19),Minimum(5,-137)