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

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

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

extreme points

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

Minimum (0, - 1)

Solution steps

f^{'} (x) = \frac{16 x}{( x ^{2} + 4 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

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

Minimum (0, - 1)

f (x, y) = 16 - 4 x^{2} - y^{2}

f (x, y) = ln (x + y - 1)

f (x, y) = y - x - 2

f (x) = x^{2} + 3 x - 8

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 15

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