What is the extreme y=(32x)/(x^2+16) ?

The extreme y=(32x)/(x^2+16) is Minimum(-4,-4),Maximum(4,4)

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extreme y=(32x)/(x^2+16)

extreme points

y = \frac{32 x}{x ^{2} + 16}

Minimum (- 4, - 4), Maximum (4, 4)

Solution steps

f^{'} (x) = \frac{32 ( - x ^{2} + 16 )}{( x ^{2} + 16 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = - 4

into

\frac{32 x}{x ^{2} + 16} : - 4

Minimum (- 4, - 4)

Plug

x = 4

into

\frac{32 x}{x ^{2} + 16} : 4

Maximum (4, 4)

Minimum (- 4, - 4), Maximum (4, 4)

y = 2 x - 7 ln (x)

f (x) = (x + 2) (x + 7)

f (x) = - 2.3

f (x) = 4 + 6 x^{2} - x^{4} - 2

f (x) = x^{3} + 12 x^{2} - 27 x + 2, - 10 \leq x \leq 0

What is the extreme y=(32x)/(x^2+16) ?
The extreme y=(32x)/(x^2+16) is Minimum(-4,-4),Maximum(4,4)