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

The extreme f(x)=((x-1)^2)/(x^2+2) is Maximum(-2, 3/2),Minimum(1,0)

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

extreme points

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

Maximum (- 2, \frac{3}{2}), Minimum (1, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

\frac{( x - 1 ) ^{2}}{x ^{2} + 2} : \frac{3}{2}

Maximum (- 2, \frac{3}{2})

Plug

x = 1

into

\frac{( x - 1 ) ^{2}}{x ^{2} + 2} : 0

Minimum (1, 0)

Maximum (- 2, \frac{3}{2}), Minimum (1, 0)

What is the extreme f(x)=((x-1)^2)/(x^2+2) ?
The extreme f(x)=((x-1)^2)/(x^2+2) is Maximum(-2, 3/2),Minimum(1,0)