What is the extreme f(x)=(x^2-2x-8)^{1/3} ?

The extreme f(x)=(x^2-2x-8)^{1/3} is Minimum(1,-9^{1/3})

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extreme f(x)=(x^2-2x-8)^{1/3}

extreme points

f (x) = (x^{2} - 2 x - 8)^{\frac{1}{3}}

Minimum (1, - 9^{\frac{1}{3}})

Solution steps

f^{'} (x) = \frac{2 x - 2}{3 ( x ^{2} - 2 x - 8 ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 1

into

(x^{2} - 2 x - 8)^{\frac{1}{3}} : - 9^{\frac{1}{3}}

Minimum (1, - 9^{\frac{1}{3}})

f (x) = x^{2} - 10 x - 7

f (x) = (4 x)^{x}, 0.05 \leq x \leq 1

f (x) = ln (x^{2} + 5 x + 9)

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

f (χ) = χ^{2} e^{- 4 χ^{2}}

What is the extreme f(x)=(x^2-2x-8)^{1/3} ?
The extreme f(x)=(x^2-2x-8)^{1/3} is Minimum(1,-9^{1/3})