What is the extreme f(x)=(7x)/(sqrt(x-4)) ?

The extreme f(x)=(7x)/(sqrt(x-4)) is Minimum(8,28)

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extreme f(x)=(7x)/(sqrt(x-4))

extreme points

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

Minimum (8, 28)

Solution steps

f^{'} (x) = \frac{7 ( x - 8 )}{2 ( x - 4 ) x - 4}

Find intervals:

Decreasing

: 4 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 8

into

\frac{7 x}{x - 4} : 28

Minimum (8, 28)

21 x^{5} + 35 x^{4}

f (x) = \frac{x + 1}{x ^{2} - 8 x - 9}

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

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

f (x, y) = (x + 1)^{2} + (y - 3)^{2}

What is the extreme f(x)=(7x)/(sqrt(x-4)) ?
The extreme f(x)=(7x)/(sqrt(x-4)) is Minimum(8,28)