What is the extreme f(x)= 1/(x^2-6x+12) ?

The extreme f(x)= 1/(x^2-6x+12) is Maximum(3, 1/3)

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extreme f(x)= 1/(x^2-6x+12)

extreme points

f (x) = \frac{1}{x ^{2} - 6 x + 12}

Maximum (3, \frac{1}{3})

Solution steps

f^{'} (x) = - \frac{2 x - 6}{( x ^{2} - 6 x + 12 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = 3

into

\frac{1}{x ^{2} - 6 x + 12} : \frac{1}{3}

Maximum (3, \frac{1}{3})

What is the extreme f(x)= 1/(x^2-6x+12) ?
The extreme f(x)= 1/(x^2-6x+12) is Maximum(3, 1/3)