What is the extreme f(x)=x^{1/5}(x+12) ?

The extreme f(x)=x^{1/5}(x+12) is Minimum(-2,-10*2^{1/5})

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extreme f(x)=x^{1/5}(x+12)

extreme points

f (x) = x^{\frac{1}{5}} (x + 12)

Minimum (- 2, - 10 \cdot 2^{\frac{1}{5}})

Solution steps

f^{'} (x) = \frac{6 ( x + 2 )}{5 x ^{\frac{4}{5}}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

x^{\frac{1}{5}} (x + 12) : - 10 \cdot 2^{\frac{1}{5}}

Minimum (- 2, - 10 \cdot 2^{\frac{1}{5}})

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f (x) = ((x - 6)^{(\frac{4}{7})}) - 3

x^{3} - 8 y^{3} + 4 x y - 8

f (x, y) = 2 x^{2} - 4 x + y^{2} - 6 y + 1

f (x) = (24 + x) (600 - 15 x)

What is the extreme f(x)=x^{1/5}(x+12) ?
The extreme f(x)=x^{1/5}(x+12) is Minimum(-2,-10*2^{1/5})