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

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

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

extreme points

(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, y) = 5 x^{2} + 5 y^{2} + 20 x - 10 y + 40

f (x) = x (27 - 41 + 2 x) (\frac{41}{2} - x)

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

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

f (x) = (x + 1)^{7} - 7 x - 3

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