What is the extreme f(x)=\sqrt[5]{x},-1<= x<= 0 ?

The extreme f(x)=\sqrt[5]{x},-1<= x<= 0 is Minimum(-1,-1),Maximum(0,0)

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extreme f(x)=\sqrt[5]{x},-1<= x<= 0

extreme points

f (x) = 5 x, - 1 \leq x \leq 0

Minimum (- 1, - 1), Maximum (0, 0)

Solution steps

Find the critical points:

x = - 1, x = 0

Domain of

5 x, - 1 \leq x \leq 0 : - 1 \leq x \leq 0

Find intervals:

Increasing

: - 1 < x < 0

Plug the extreme point

x = - 1

into

5 x \Rightarrow y = - 1

Minimum (- 1, - 1)

Plug the extreme point

x = 0

into

5 x \Rightarrow y = 0

Maximum (0, 0)

Minimum (- 1, - 1), Maximum (0, 0)

f (x) = x^{2} = 4 a y

f (x) = (\frac{5 ^{2} - x ^{2}}{x})^{2}

\frac{1}{7 4 + x}

3 x + 1, 3 \leq x \leq 6

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

What is the extreme f(x)=\sqrt[5]{x},-1<= x<= 0 ?
The extreme f(x)=\sqrt[5]{x},-1<= x<= 0 is Minimum(-1,-1),Maximum(0,0)