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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

x = 0

Domain of

x^{\frac{3}{2}}, (- 1, 2) : 0 \leq x < 2

Find intervals:

Increasing

: 0 < x < 2

Plug the extreme point

x = 0

into

x^{\frac{3}{2}} \Rightarrow y = 0

Minimum (0, 0)

f (x) = e^{- 3 x^{2}}

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

f (x, y) = e^{5 x y}

f (x, y) = 2 x - y

9 x^{2} ln (x)

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