What is the extreme f(x)=(80x)/(x^2+x+4) ?

The extreme f(x)=(80x)/(x^2+x+4) is Minimum(-2,-80/3),Maximum(2,16)

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extreme f(x)=(80x)/(x^2+x+4)

extreme points

f (x) = \frac{80 x}{x ^{2} + x + 4}

Minimum (- 2, - \frac{80}{3}), Maximum (2, 16)

Solution steps

f^{'} (x) = \frac{80 ( - x ^{2} + 4 )}{( x ^{2} + x + 4 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

\frac{80 x}{x ^{2} + x + 4} : - \frac{80}{3}

Minimum (- 2, - \frac{80}{3})

Plug

x = 2

into

\frac{80 x}{x ^{2} + x + 4} : 16

Maximum (2, 16)

Minimum (- 2, - \frac{80}{3}), Maximum (2, 16)

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

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

f (x) = - 5 x^{3} + x^{2} + x - 7

f (x) = x^{\frac{4}{7}} (3 x + 11)

f (x) = - 0.2 x^{2} + 1.6 x + 98.6

What is the extreme f(x)=(80x)/(x^2+x+4) ?
The extreme f(x)=(80x)/(x^2+x+4) is Minimum(-2,-80/3),Maximum(2,16)