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

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

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

extreme points

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

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

Solution steps

f^{'} (t) = \frac{1 - t ^{2}}{( 1 + t ^{2} ) ^{2}}

Find intervals:

Decreasing

: - \infty < t < - 1,

Increasing

: - 1 < t < 1,

Decreasing

: 1 < t < \infty

Plug

t = - 1

into

\frac{t}{1 + t ^{2}} : - \frac{1}{2}

Minimum (- 1, - \frac{1}{2})

Plug

t = 1

into

\frac{t}{1 + t ^{2}} : \frac{1}{2}

Maximum (1, \frac{1}{2})

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

f (x) = \frac{5}{3} x^{3} - 25 x

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

f (x) = - 4 x^{4} + 120 x^{2} + 120

\frac{x ^{2} - 63}{x - 8}

f (x) = 9 x^{3} + 2 1^{2} + 8 x - 3

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