What is the extreme f(x)=x^{8/7}+8x^{1/7} ?

The extreme f(x)=x^{8/7}+8x^{1/7} is Minimum(-1,-7)

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extreme f(x)=x^{8/7}+8x^{1/7}

extreme points

f (x) = x^{\frac{8}{7}} + 8 x^{\frac{1}{7}}

Minimum (- 1, - 7)

Solution steps

f^{'} (x) = \frac{8 x + 8}{7 x ^{\frac{6}{7}}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

x^{\frac{8}{7}} + 8 x^{\frac{1}{7}} : - 7

Minimum (- 1, - 7)

f (x) = \frac{x}{x - 6}

f (x) = - x^{2} - 5 x + 2

f (x, y) = x^{2} + y^{2} + 8 x - 12 y - 9

f (x, y) = 2 x y + 3 x + 4 y

f (x, y) = 420 x^{2} - 80 x + 150 y^{2} - 75 y + x y

What is the extreme f(x)=x^{8/7}+8x^{1/7} ?
The extreme f(x)=x^{8/7}+8x^{1/7} is Minimum(-1,-7)