What is the extreme f(x)=(x^2-4)/(x^2+1) ?

The extreme f(x)=(x^2-4)/(x^2+1) is Minimum(0,-4)

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extreme f(x)=(x^2-4)/(x^2+1)

extreme points

f (x) = \frac{x ^{2} - 4}{x ^{2} + 1}

Minimum (0, - 4)

Solution steps

f^{'} (x) = \frac{10 x}{( x ^{2} + 1 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{x ^{2} - 4}{x ^{2} + 1} : - 4

Minimum (0, - 4)

f (x) = x^{3} - 6 x^{2} + 8, - 4 \leq x \leq 1

f (x) = 6 - 3 x^{2} - x^{3}

f (x) = x^{\frac{8}{3}}, - 1 \leq x \leq 8

12 x^{5} + 120 x^{4} - 300 x^{3}

f (x) = x + 2 cos (x), 0 \leq x \leq 2 π

What is the extreme f(x)=(x^2-4)/(x^2+1) ?
The extreme f(x)=(x^2-4)/(x^2+1) is Minimum(0,-4)