What is the extreme f(x)=(16t)/(t^2+64) ?

The extreme f(x)=(16t)/(t^2+64) is Minimum(-8,-1),Maximum(8,1)

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extreme f(x)=(16t)/(t^2+64)

extreme points

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

Minimum (- 8, - 1), Maximum (8, 1)

Solution steps

f^{'} (t) = \frac{16 ( - t ^{2} + 64 )}{( t ^{2} + 64 ) ^{2}}

Find intervals:

Decreasing

: - \infty < t < - 8,

Increasing

: - 8 < t < 8,

Decreasing

: 8 < t < \infty

Plug

t = - 8

into

\frac{16 t}{t ^{2} + 64} : - 1

Minimum (- 8, - 1)

Plug

t = 8

into

\frac{16 t}{t ^{2} + 64} : 1

Maximum (8, 1)

Minimum (- 8, - 1), Maximum (8, 1)

f (x) = 7 + x - x^{2} - ln ((x + 3)^{3})

(x^{2} - 1)^{3}, - 1 \leq x \leq 3

f (x) = 6 x^{2} - 7 x - 11

f (x, y) = (x^{2} + x y + y^{2} - 19 y + 120)

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

What is the extreme f(x)=(16t)/(t^2+64) ?
The extreme f(x)=(16t)/(t^2+64) is Minimum(-8,-1),Maximum(8,1)