What is the extreme f(x)= 4/(x-2),2<x<= 6 ?

The extreme f(x)= 4/(x-2),2<x<= 6 is Minimum(6,1)

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extreme f(x)= 4/(x-2),2<x<= 6

extreme points

f (x) = \frac{4}{x - 2}, 2 < x \leq 6

Minimum (6, 1)

Solution steps

Find the critical points:

x = 6

Domain of

\frac{4}{x - 2}, 2 < x \leq 6 : 2 < x \leq 6

Find intervals:

Decreasing

: 2 < x < 6

Plug the extreme point

x = 6

into

\frac{4}{x - 2} \Rightarrow y = 1

Minimum (6, 1)

f (x) = 3 x + 2 y + 7

f (x) = 2 x^{3} + 6 x^{2} - 48 x

f (x) = \frac{1}{2} p (x, y) - cy = 0

f (x) = 3 x + 2 y - 1

y = \frac{1}{2} x + \frac{1}{2} z - \frac{3}{2}

What is the extreme f(x)= 4/(x-2),2<x<= 6 ?
The extreme f(x)= 4/(x-2),2<x<= 6 is Minimum(6,1)