What is the extreme f(x)=x(1-x^2)^{1/2} ?

The extreme f(x)=x(1-x^2)^{1/2} is

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extreme f(x)=x(1-x^2)^{1/2}

extreme points

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

Maximum (- 1, 0), Minimum (- \frac{1}{2}, - \frac{1}{2 ^{\frac{1}{2}} 2}), Maximum (\frac{1}{2}, \frac{1}{2 ^{\frac{1}{2}} 2}), Minimum (1, 0)

Solution steps

Find the critical points:

x = - 1, x = - \frac{1}{2}, x = \frac{1}{2}, x = 1

Domain of

x (1 - x^{2})^{\frac{1}{2}} : - 1 \leq x \leq 1

Find intervals:

Decreasing

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

Increasing

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

Decreasing

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

Plug the extreme point

x = - 1

into

x (1 - x^{2})^{\frac{1}{2}} \Rightarrow y = 0

Maximum (- 1, 0)

Plug the extreme point

x = - \frac{1}{2}

into

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

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

Plug the extreme point

x = \frac{1}{2}

into

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

Maximum (\frac{1}{2}, \frac{1}{2 ^{\frac{1}{2}} 2})

Plug the extreme point

x = 1

into

x (1 - x^{2})^{\frac{1}{2}} \Rightarrow y = 0

Minimum (1, 0)

Maximum (- 1, 0), Minimum (- \frac{1}{2}, - \frac{1}{2 ^{\frac{1}{2}} 2}), Maximum (\frac{1}{2}, \frac{1}{2 ^{\frac{1}{2}} 2}), Minimum (1, 0)

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

f (x) = x^{2} + 8 x + 4

\frac{8}{x x ^{2} - 4}

f (x) = x + x + 7

- x + 5