What is the extreme f(x)=x^{1/5}(x+24) ?

The extreme f(x)=x^{1/5}(x+24) is Minimum(-4,-20*4^{1/5})

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extreme f(x)=x^{1/5}(x+24)

extreme points

f (x) = x^{\frac{1}{5}} (x + 24)

Minimum (- 4, - 20 \cdot 4^{\frac{1}{5}})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 4

into

x^{\frac{1}{5}} (x + 24) : - 20 \cdot 4^{\frac{1}{5}}

Minimum (- 4, - 20 \cdot 4^{\frac{1}{5}})

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

\frac{6 ∣ x ∣}{∣ x - 1 ∣}

f (x) = \frac{1}{3} x^{3} - \frac{3}{2} x^{2} - 18 x + \frac{1}{3}

\frac{2 x - 6}{x + 3}

f (x, y) = x y + x^{2} + y^{2} + x - y + 17

What is the extreme f(x)=x^{1/5}(x+24) ?
The extreme f(x)=x^{1/5}(x+24) is Minimum(-4,-20*4^{1/5})