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

The extreme f(x)=(2-x^2)^{3/2} is Minimum(-sqrt(2),0),Maximum(0,2*2^{1/2}),Minimum(sqrt(2),0)

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

extreme

Minimum (- 2, 0), Maximum (0, 2 \cdot 2^{\frac{1}{2}}), Minimum (2, 0)

Solution steps

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

Find intervals:

Increasing

: - 2^{\frac{1}{2}} < x < 0,

Decreasing

: 0 < x < 2^{\frac{1}{2}}

Plug

x = - 2

into

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

Minimum (- 2, 0)

Plug

x = 0

into

(2 - x^{2})^{\frac{3}{2}} : 2 \cdot 2^{\frac{1}{2}}

Maximum (0, 2 \cdot 2^{\frac{1}{2}})

Plug

x = 2

into

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

Minimum (2, 0)

Minimum (- 2, 0), Maximum (0, 2 \cdot 2^{\frac{1}{2}}), Minimum (2, 0)

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