What is the extreme f(x)=(x^2-16)/(x^2+16) ?

The extreme f(x)=(x^2-16)/(x^2+16) is Minimum(0,-1)

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extreme f(x)=(x^2-16)/(x^2+16)

extreme points

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

Minimum (0, - 1)

Solution steps

f^{'} (x) = \frac{64 x}{( x ^{2} + 16 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{x ^{2} - 16}{x ^{2} + 16} : - 1

Minimum (0, - 1)

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

f (x) = - x^{4} + 4 x^{3} + 8 x^{2}

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

(x^{2} + y^{2}) e^{y^{2} - x^{2}}

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

What is the extreme f(x)=(x^2-16)/(x^2+16) ?
The extreme f(x)=(x^2-16)/(x^2+16) is Minimum(0,-1)