What is the extreme f(x)=(x^2+13)^{1/3} ?

The extreme f(x)=(x^2+13)^{1/3} is Minimum(0,13^{1/3})

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extreme f(x)=(x^2+13)^{1/3}

extreme points

f (x) = (x^{2} + 13)^{\frac{1}{3}}

Minimum (0, 1 3^{\frac{1}{3}})

Solution steps

f^{'} (x) = \frac{2 x}{3 ( x ^{2} + 13 ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

(x^{2} + 13)^{\frac{1}{3}} : 1 3^{\frac{1}{3}}

Minimum (0, 1 3^{\frac{1}{3}})

f (x) = \frac{x}{x - 2}, 4 \leq x \leq 6

\frac{x}{x ^{2} - 16}

f (x) = 4 sin (2 x)

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

f (x, y) = x^{2} - y^{2} - 20 x + 2 y + 8

What is the extreme f(x)=(x^2+13)^{1/3} ?
The extreme f(x)=(x^2+13)^{1/3} is Minimum(0,13^{1/3})