What is the extreme (x^2-4)^{1/7},-2<= x<= 3 ?

The extreme (x^2-4)^{1/7},-2<= x<= 3 is Maximum(-2,0),Minimum(0,-4^{1/7}),Maximum(3,5^{1/7})

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extreme (x^2-4)^{1/7},-2<= x<= 3

extreme points

(x^{2} - 4)^{\frac{1}{7}}, - 2 \leq x \leq 3

Maximum (- 2, 0), Minimum (0, - 4^{\frac{1}{7}}), Maximum (3, 5^{\frac{1}{7}})

Solution steps

Find the critical points:

x = - 2, x = 0, x = 2, x = 3

Domain of

(x^{2} - 4)^{\frac{1}{7}}, - 2 \leq x \leq 3 : - 2 \leq x \leq 3

Find intervals:

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < 3

Plug the extreme point

x = - 2

into

(x^{2} - 4)^{\frac{1}{7}} \Rightarrow y = 0

Maximum (- 2, 0)

Plug the extreme point

x = 0

into

(x^{2} - 4)^{\frac{1}{7}} \Rightarrow y = - 4^{\frac{1}{7}}

Minimum (0, - 4^{\frac{1}{7}})

Plug the extreme point

x = 3

into

(x^{2} - 4)^{\frac{1}{7}} \Rightarrow y = 5^{\frac{1}{7}}

Maximum (3, 5^{\frac{1}{7}})

Maximum (- 2, 0), Minimum (0, - 4^{\frac{1}{7}}), Maximum (3, 5^{\frac{1}{7}})

f (x) = \frac{3}{2} t^{4} - t^{6}

f (x, y) = - x^{2} - y^{3} + 3 y^{2} - 2 x + 45 y + 1

y = - 12 x^{5} - 45 x^{4} + 20 x^{3} - 1

f (x) = x^{3} - 2 x - 4 x + 4 (0.3)

x^{2} + x^{- 2}

What is the extreme (x^2-4)^{1/7},-2<= x<= 3 ?
The extreme (x^2-4)^{1/7},-2<= x<= 3 is Maximum(-2,0),Minimum(0,-4^{1/7}),Maximum(3,5^{1/7})