What is the extreme f(x)= 1/((x^2-2x+7)) ?

The extreme f(x)= 1/((x^2-2x+7)) is Maximum(1, 1/6)

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extreme f(x)= 1/((x^2-2x+7))

extreme points

f (x) = \frac{1}{( x ^{2} - 2 x + 7 )}

Maximum (1, \frac{1}{6})

Solution steps

f^{'} (x) = - \frac{2 x - 2}{( x ^{2} - 2 x + 7 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 1

into

\frac{1}{x ^{2} - 2 x + 7} : \frac{1}{6}

Maximum (1, \frac{1}{6})

f (x) = 3 x^{2} - 27

x^{2} - 14 x + 3

f (x) = 39 + 6 x + \frac{54}{x}

\frac{e ^{x}}{x ^{4}}

F (x, y) = x^{2} + x y + y^{2} + 3 x - 3 y + 4

What is the extreme f(x)= 1/((x^2-2x+7)) ?
The extreme f(x)= 1/((x^2-2x+7)) is Maximum(1, 1/6)