What is the extreme f(x)=(4x-12)/((x-2)^2) ?

The extreme f(x)=(4x-12)/((x-2)^2) is Maximum(4,1)

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extreme f(x)=(4x-12)/((x-2)^2)

extreme points

f (x) = \frac{4 x - 12}{( x - 2 ) ^{2}}

Maximum (4, 1)

Solution steps

f^{'} (x) = \frac{4 ( - x + 4 )}{( x - 2 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < 2,

Increasing

: 2 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 4

into

\frac{4 x - 12}{( x - 2 ) ^{2}} : 1

Maximum (4, 1)

f (x) = \frac{I n x}{x}

K (r, s) = 5 r - 9 s

x^{\frac{2}{3}} (x^{2} - 4)

f (x) = e^{k x}

x^{3} - 3 x^{2} - 9 x + 230

What is the extreme f(x)=(4x-12)/((x-2)^2) ?
The extreme f(x)=(4x-12)/((x-2)^2) is Maximum(4,1)