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

The extreme f(x)=(x^2+3x-1)^{1/3} is Minimum(-3/2 ,-(13/4)^{1/3})

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

extreme points

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

Minimum (- \frac{3}{2}, - (\frac{13}{4})^{\frac{1}{3}})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < \frac{- 3 - 1 3 ^{\frac{1}{2}}}{2},

Decreasing

: \frac{- 3 - 1 3 ^{\frac{1}{2}}}{2} < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < \frac{- 3 + 1 3 ^{\frac{1}{2}}}{2},

Increasing

: \frac{- 3 + 1 3 ^{\frac{1}{2}}}{2} < x < \infty

Plug

x = - \frac{3}{2}

into

(x^{2} + 3 x - 1)^{\frac{1}{3}} : - (\frac{13}{4})^{\frac{1}{3}}

Minimum (- \frac{3}{2}, - (\frac{13}{4})^{\frac{1}{3}})

f (x) = x^{3} - 4 x^{2} - x + 1

f (x, y) = \frac{20000}{3 + x ^{2} + y ^{2}}

f (x) = 2 x^{3} - 15 x^{2} + 24 x, 0 \leq x \leq 5

y = \frac{x}{x ^{2} + 64}

f (x) = 3 x^{4} - 324 x, 0 \leq x \leq 4

What is the extreme f(x)=(x^2+3x-1)^{1/3} ?
The extreme f(x)=(x^2+3x-1)^{1/3} is Minimum(-3/2 ,-(13/4)^{1/3})