What is the extreme f(x)=xsqrt(13-x^2) ?

The extreme f(x)=xsqrt(13-x^2) is

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extreme f(x)=xsqrt(13-x^2)

extreme points

f (x) = x 13 - x^{2}

Maximum (- 13, 0), Minimum (- \frac{13}{2}, - \frac{13}{2}), Maximum (\frac{13}{2}, \frac{13}{2}), Minimum (13, 0)

Solution steps

Find the critical points:

x = - 13, x = - \frac{13}{2}, x = \frac{13}{2}, x = 13

Domain of

x 13 - x^{2} : - 13 \leq x \leq 13

Find intervals:

Decreasing

: - 13 < x < - \frac{13}{2},

Increasing

: - \frac{13}{2} < x < \frac{13}{2},

Decreasing

: \frac{13}{2} < x < 13

Plug the extreme point

x = - 13

into

x 13 - x^{2} \Rightarrow y = 0

Maximum (- 13, 0)

Plug the extreme point

x = - \frac{13}{2}

into

x 13 - x^{2} \Rightarrow y = - \frac{13}{2}

Minimum (- \frac{13}{2}, - \frac{13}{2})

Plug the extreme point

x = \frac{13}{2}

into

x 13 - x^{2} \Rightarrow y = \frac{13}{2}

Maximum (\frac{13}{2}, \frac{13}{2})

Plug the extreme point

x = 13

into

x 13 - x^{2} \Rightarrow y = 0

Minimum (13, 0)

Maximum (- 13, 0), Minimum (- \frac{13}{2}, - \frac{13}{2}), Maximum (\frac{13}{2}, \frac{13}{2}), Minimum (13, 0)

x^{4} - 8 x^{2} - 10

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f (r, h) = 2 π r h + 2 π r^{2}

g (t) = u (t)