What is the extreme y= 1/5 x^5-8/3 x^3+16x ?

The extreme y= 1/5 x^5-8/3 x^3+16x is Saddle(-2,-256/15),Saddle(2, 256/15)

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extreme y= 1/5 x^5-8/3 x^3+16x

extreme points

y = \frac{1}{5} x^{5} - \frac{8}{3} x^{3} + 16 x

Saddle (- 2, - \frac{256}{15}), Saddle (2, \frac{256}{15})

Solution steps

Find the critical points:

x = - 2, x = 2

Domain of

\frac{1}{5} x^{5} - \frac{8}{3} x^{3} + 16 x : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = - 2

into

\frac{1}{5} x^{5} - \frac{8}{3} x^{3} + 16 x \Rightarrow y = - \frac{256}{15}

Saddle (- 2, - \frac{256}{15})

Plug the extreme point

x = 2

into

\frac{1}{5} x^{5} - \frac{8}{3} x^{3} + 16 x \Rightarrow y = \frac{256}{15}

Saddle (2, \frac{256}{15})

Saddle (- 2, - \frac{256}{15}), Saddle (2, \frac{256}{15})

f (x) = \frac{( x ^{2} - 7 )}{x - 4}

\frac{x ^{2} + 1}{x ^{2} - 1}

f (x) = \frac{x ^{2} - 12}{x - 4}

(\frac{x ^{2}}{4} - \frac{x}{2} - \frac{1}{4})^{2}

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 6

What is the extreme y= 1/5 x^5-8/3 x^3+16x ?
The extreme y= 1/5 x^5-8/3 x^3+16x is Saddle(-2,-256/15),Saddle(2, 256/15)