What is the extreme (t^2-4)^{2/3} ?

The extreme (t^2-4)^{2/3} is Minimum(-2,0),Maximum(0,(-4)^{2/3}),Minimum(2,0)

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extreme (t^2-4)^{2/3}

extreme points

(t^{2} - 4)^{\frac{2}{3}}

Minimum (- 2, 0), Maximum (0, (- 4)^{\frac{2}{3}}), Minimum (2, 0)

Solution steps

f^{'} (t) = \frac{4 t}{3 ( t ^{2} - 4 ) ^{\frac{1}{3}}}

Find intervals:

Decreasing

: - \infty < t < - 2,

Increasing

: - 2 < t < 0,

Decreasing

: 0 < t < 2,

Increasing

: 2 < t < \infty

Plug

t = - 2

into

(t^{2} - 4)^{\frac{2}{3}} : 0

Minimum (- 2, 0)

Plug

t = 0

into

(t^{2} - 4)^{\frac{2}{3}} : (- 4)^{\frac{2}{3}}

Maximum (0, (- 4)^{\frac{2}{3}})

Plug

t = 2

into

(t^{2} - 4)^{\frac{2}{3}} : 0

Minimum (2, 0)

Minimum (- 2, 0), Maximum (0, (- 4)^{\frac{2}{3}}), Minimum (2, 0)

f (x) = 11 x^{2} - 8

y = 2 x^{3} - 5

\frac{2}{x - 5}

f (x) = sin^{2} (x)

What is the extreme (t^2-4)^{2/3} ?
The extreme (t^2-4)^{2/3} is Minimum(-2,0),Maximum(0,(-4)^{2/3}),Minimum(2,0)