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

The extreme f(x)=(x-3)^{4/3},-7<= x<= 7 is Maximum(-7,10*10^{1/3}),Minimum(3,0),Maximum(7,4*4^{1/3})

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

extreme points

f (x) = (x - 3)^{\frac{4}{3}}, - 7 \leq x \leq 7

Maximum (- 7, 10 \cdot 1 0^{\frac{1}{3}}), Minimum (3, 0), Maximum (7, 4 \cdot 4^{\frac{1}{3}})

Solution steps

Find the critical points:

x = - 7, x = 3, x = 7

Domain of

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

Find intervals:

Decreasing

: - 7 < x < 3,

Increasing

: 3 < x < 7

Plug the extreme point

x = - 7

into

(x - 3)^{\frac{4}{3}} \Rightarrow y = 101 0^{\frac{1}{3}}

Maximum (- 7, 10 \cdot 1 0^{\frac{1}{3}})

Plug the extreme point

x = 3

into

(x - 3)^{\frac{4}{3}} \Rightarrow y = 0

Minimum (3, 0)

Plug the extreme point

x = 7

into

(x - 3)^{\frac{4}{3}} \Rightarrow y = 4 4^{\frac{1}{3}}

Maximum (7, 4 \cdot 4^{\frac{1}{3}})

Maximum (- 7, 10 \cdot 1 0^{\frac{1}{3}}), Minimum (3, 0), Maximum (7, 4 \cdot 4^{\frac{1}{3}})

2 x^{3} - 2 x^{2} - x - 9

y = \frac{x ^{2}}{2} - 3 x + \frac{7}{2}

P (x, y) = - 4 x + 5 y - x^{2} + 2 x y - 7 y^{2} + 9

f (x) = 2 x^{2} + 5 (7 - x)^{2}

f (x) = 4 x^{3} - 3 x^{4} + 5

What is the extreme f(x)=(x-3)^{4/3},-7<= x<= 7 ?
The extreme f(x)=(x-3)^{4/3},-7<= x<= 7 is Maximum(-7,10*10^{1/3}),Minimum(3,0),Maximum(7,4*4^{1/3})