What is the extreme f(x)=(x^2-4)^{3/4} ?

The extreme f(x)=(x^2-4)^{3/4} is Minimum(-2,0),Minimum(2,0)

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extreme f(x)=(x^2-4)^{3/4}

extreme points

f (x) = (x^{2} - 4)^{\frac{3}{4}}

Minimum (- 2, 0), Minimum (2, 0)

Solution steps

f^{'} (x) = \frac{3 x}{2 ( x ^{2} - 4 ) ^{\frac{1}{4}}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

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

Minimum (- 2, 0)

Plug

x = 2

into

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

Minimum (2, 0)

Minimum (- 2, 0), Minimum (2, 0)

x^{4} - 8 x^{3} + 7 x^{2} + 6 x - 4

f (x, y) = 5 - (x - 6)^{2} - (y + 2)^{2}

2 (x + 5)

r (x)

f (x) = \frac{2 x ( x ^{2} - 6 x )}{( x - 2 ) ^{2}}

What is the extreme f(x)=(x^2-4)^{3/4} ?
The extreme f(x)=(x^2-4)^{3/4} is Minimum(-2,0),Minimum(2,0)