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

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

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

extreme points

\frac{x ^{2} - 4 x}{x ^{2} - 4 x - 12}

Maximum (2, \frac{1}{4})

Solution steps

f^{'} (x) = \frac{- 24 x + 48}{( x ^{2} - 4 x - 12 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < 6,

Decreasing

: 6 < x < \infty

Plug

x = 2

into

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

Maximum (2, \frac{1}{4})

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