What is the extreme f(x)=(0.2t)/(t^2+59) ?

The extreme f(x)=(0.2t)/(t^2+59) is Minimum(-sqrt(59),-0.01301…),Maximum(sqrt(59),0.01301…)

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extreme f(x)=(0.2t)/(t^2+59)

extreme points

f (x) = \frac{0.2 t}{t ^{2} + 59}

Minimum (- 59, - 0.01301 \dots), Maximum (59, 0.01301 \dots)

Solution steps

f^{'} (t) = \frac{0.2 ( - t ^{2} + 59 )}{( t ^{2} + 59 ) ^{2}}

Find intervals:

Decreasing

: - \infty < t < - 59,

Increasing

: - 59 < t < 59,

Decreasing

: 59 < t < \infty

Plug

t = - 59

into

\frac{0.2 t}{t ^{2} + 59} : - 0.01301 \dots

Minimum (- 59, - 0.01301 \dots)

Plug

t = 59

into

\frac{0.2 t}{t ^{2} + 59} : 0.01301 \dots

Maximum (59, 0.01301 \dots)

Minimum (- 59, - 0.01301 \dots), Maximum (59, 0.01301 \dots)

\frac{18}{x} + 3 x^{2}

f (x) = 3 x^{2} - 4 x + 12

f (x) = x (- \frac{2 x}{3} + 4)

f (x, y) = 2 x^{2} + 3 y^{2} + 2 x y - 6 x - 8 y + 11

f (x, y) = x^{2} - 2 x y + \frac{1}{3} y^{3} - 4 y

What is the extreme f(x)=(0.2t)/(t^2+59) ?
The extreme f(x)=(0.2t)/(t^2+59) is Minimum(-sqrt(59),-0.01301…),Maximum(sqrt(59),0.01301…)