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

The extreme f(x)=(-x^2)/(x^2-2x+8) is Maximum(0,0),Minimum(8,-8/7)

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

extreme points

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

Maximum (0, 0), Minimum (8, - \frac{8}{7})

Solution steps

f^{'} (x) = - \frac{- 2 x ^{2} + 16 x}{( x ^{2} - 2 x + 8 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 0

into

\frac{- x ^{2}}{x ^{2} - 2 x + 8} : 0

Maximum (0, 0)

Plug

x = 8

into

\frac{- x ^{2}}{x ^{2} - 2 x + 8} : - \frac{8}{7}

Minimum (8, - \frac{8}{7})

Maximum (0, 0), Minimum (8, - \frac{8}{7})

What is the extreme f(x)=(-x^2)/(x^2-2x+8) ?
The extreme f(x)=(-x^2)/(x^2-2x+8) is Maximum(0,0),Minimum(8,-8/7)