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

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

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

extreme

Minimum (- 2, \frac{1}{2}), Maximum (1, - 1)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = - 2

into

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

Minimum (- 2, \frac{1}{2})

Plug

x = 1

into

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

Maximum (1, - 1)

Minimum (- 2, \frac{1}{2}), Maximum (1, - 1)

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