What is the extreme f(x)=x-16/3 x^3 ?

The extreme f(x)=x-16/3 x^3 is Minimum(-1/4 ,-1/6),Maximum(1/4 , 1/6)

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extreme f(x)=x-16/3 x^3

extreme points

f (x) = x - \frac{16}{3} x^{3}

Minimum (- \frac{1}{4}, - \frac{1}{6}), Maximum (\frac{1}{4}, \frac{1}{6})

Solution steps

f^{'} (x) = 1 - 16 x^{2}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{4},

Increasing

: - \frac{1}{4} < x < \frac{1}{4},

Decreasing

: \frac{1}{4} < x < \infty

Plug

x = - \frac{1}{4}

into

x - \frac{16}{3} x^{3} : - \frac{1}{6}

Minimum (- \frac{1}{4}, - \frac{1}{6})

Plug

x = \frac{1}{4}

into

x - \frac{16}{3} x^{3} : \frac{1}{6}

Maximum (\frac{1}{4}, \frac{1}{6})

Minimum (- \frac{1}{4}, - \frac{1}{6}), Maximum (\frac{1}{4}, \frac{1}{6})

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

f (x) = x^{4} - 8 x^{3} + 16 x^{2} - 10

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

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

f (x) = (x^{2} - 1)^{\frac{1}{3}}

What is the extreme f(x)=x-16/3 x^3 ?
The extreme f(x)=x-16/3 x^3 is Minimum(-1/4 ,-1/6),Maximum(1/4 , 1/6)