What is the extreme f(x)=(60t)/(t^2+36) ?

The extreme f(x)=(60t)/(t^2+36) is Minimum(-6,-5),Maximum(6,5)

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extreme f(x)=(60t)/(t^2+36)

extreme points

f (x) = \frac{60 t}{t ^{2} + 36}

Minimum (- 6, - 5), Maximum (6, 5)

Solution steps

f^{'} (t) = \frac{60 ( - t ^{2} + 36 )}{( t ^{2} + 36 ) ^{2}}

Find intervals:

Decreasing

: - \infty < t < - 6,

Increasing

: - 6 < t < 6,

Decreasing

: 6 < t < \infty

Plug

t = - 6

into

\frac{60 t}{t ^{2} + 36} : - 5

Minimum (- 6, - 5)

Plug

t = 6

into

\frac{60 t}{t ^{2} + 36} : 5

Maximum (6, 5)

Minimum (- 6, - 5), Maximum (6, 5)

f (x) = \frac{3}{x} + 9

f (x) = - x^{5} - 3 x^{4} + 2 x^{2}

f (x) = \frac{cos ( x )}{x}

f (x) = \frac{5 + 4 x}{x + 3}

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

What is the extreme f(x)=(60t)/(t^2+36) ?
The extreme f(x)=(60t)/(t^2+36) is Minimum(-6,-5),Maximum(6,5)