What is the extreme f(x)=((x-6)^{(4/7)})-3 ?

The extreme f(x)=((x-6)^{(4/7)})-3 is Minimum(6,-3)

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extreme f(x)=((x-6)^{(4/7)})-3

extreme points

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

Minimum (6, - 3)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 6

into

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

Minimum (6, - 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)

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

f (x) = cos (x) - 6 x, 0 \leq x \leq 4 π

What is the extreme f(x)=((x-6)^{(4/7)})-3 ?
The extreme f(x)=((x-6)^{(4/7)})-3 is Minimum(6,-3)