What is the extreme f(x)=(t^2-36)^{1/3} ?

The extreme f(x)=(t^2-36)^{1/3} is Minimum(0,-36^{1/3})

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extreme f(x)=(t^2-36)^{1/3}

extreme points

f (x) = (t^{2} - 36)^{\frac{1}{3}}

Minimum (0, - 3 6^{\frac{1}{3}})

Solution steps

f^{'} (t) = \frac{2 t}{3 ( t ^{2} - 36 ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < t < - 6,

Decreasing

: - 6 < t < 0,

Increasing

: 0 < t < 6,

Increasing

: 6 < t < \infty

Plug

t = 0

into

(t^{2} - 36)^{\frac{1}{3}} : - 3 6^{\frac{1}{3}}

Minimum (0, - 3 6^{\frac{1}{3}})

\frac{7}{( x - 4 ) ^{3}}

f (x) = x^{7}

5 x + 6 y = 5

f (x) = \frac{x ^{2} - 4 x}{x ^{2} - 16}

What is the extreme f(x)=(t^2-36)^{1/3} ?
The extreme f(x)=(t^2-36)^{1/3} is Minimum(0,-36^{1/3})