What is the extreme f(x)= 1/(x^2-8x+18) ?

The extreme f(x)= 1/(x^2-8x+18) is Maximum(4, 1/2)

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extreme f(x)= 1/(x^2-8x+18)

extreme points

f (x) = \frac{1}{x ^{2} - 8 x + 18}

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

Solution steps

f^{'} (x) = - \frac{2 x - 8}{( x ^{2} - 8 x + 18 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 4

into

\frac{1}{x ^{2} - 8 x + 18} : \frac{1}{2}

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

f (x) = (x + 7)^{2}

2 x^{2} - x - 1

f (x) = - 3 x^{2} + 5 x

f (x) = x^{2} + 1

g (x) = 2 x + 12

What is the extreme f(x)= 1/(x^2-8x+18) ?
The extreme f(x)= 1/(x^2-8x+18) is Maximum(4, 1/2)