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

The extreme f(x)=x^5-x^3+8,-1<= x<= 1 is

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

extreme points

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

Minimum (- 1, 8), Maximum (- \frac{3}{5}, - (\frac{3}{5})^{\frac{5}{2}} + (\frac{3}{5})^{\frac{3}{2}} + 8), Saddle (0, 8), Minimum (\frac{3}{5}, (\frac{3}{5})^{\frac{5}{2}} - (\frac{3}{5})^{\frac{3}{2}} + 8), Maximum (1, 8)

Solution steps

Find the critical points:

x = - 1, x = - \frac{3}{5}, x = 0, x = \frac{3}{5}, x = 1

Domain of

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

Find intervals:

Increasing

: - 1 < x < - \frac{3}{5},

Decreasing

: - \frac{3}{5} < x < 0,

Decreasing

: 0 < x < \frac{3}{5},

Increasing

: \frac{3}{5} < x < 1

Plug the extreme point

x = - 1

into

x^{5} - x^{3} + 8 \Rightarrow y = 8

Minimum (- 1, 8)

Plug the extreme point

x = - \frac{3}{5}

into

x^{5} - x^{3} + 8 \Rightarrow y = - (\frac{3}{5})^{\frac{5}{2}} + (\frac{3}{5})^{\frac{3}{2}} + 8

Maximum (- \frac{3}{5}, - (\frac{3}{5})^{\frac{5}{2}} + (\frac{3}{5})^{\frac{3}{2}} + 8)

Plug the extreme point

x = 0

into

x^{5} - x^{3} + 8 \Rightarrow y = 8

Saddle (0, 8)

Plug the extreme point

x = \frac{3}{5}

into

x^{5} - x^{3} + 8 \Rightarrow y = (\frac{3}{5})^{\frac{5}{2}} - (\frac{3}{5})^{\frac{3}{2}} + 8

Minimum (\frac{3}{5}, (\frac{3}{5})^{\frac{5}{2}} - (\frac{3}{5})^{\frac{3}{2}} + 8)

Plug the extreme point

x = 1

into

x^{5} - x^{3} + 8 \Rightarrow y = 8

Maximum (1, 8)

Minimum (- 1, 8), Maximum (- \frac{3}{5}, - (\frac{3}{5})^{\frac{5}{2}} + (\frac{3}{5})^{\frac{3}{2}} + 8), Saddle (0, 8), Minimum (\frac{3}{5}, (\frac{3}{5})^{\frac{5}{2}} - (\frac{3}{5})^{\frac{3}{2}} + 8), Maximum (1, 8)

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

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

f (x) = 12 x + 6

f (x) = x (23 - 47 + 2 x) (\frac{47}{2} - x)

f (x) = 3 x + 2 y

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