What is the extreme (810+100x+0.1x^2)/x ?

The extreme (810+100x+0.1x^2)/x is Maximum(-90,82),Minimum(90,118)

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extreme (810+100x+0.1x^2)/x

extreme points

\frac{810 + 100 x + 0.1 x ^{2}}{x}

Maximum (- 90, 82), Minimum (90, 118)

Solution steps

f^{'} (x) = \frac{0.1 x ^{2} - 810}{x ^{2}}

Find intervals:

Increasing

: - \infty < x < - 90,

Decreasing

: - 90 < x < 0,

Decreasing

: 0 < x < 90,

Increasing

: 90 < x < \infty

Plug

x = - 90

into

\frac{810 + 100 x + 0.1 x ^{2}}{x} : 82

Maximum (- 90, 82)

Plug

x = 90

into

\frac{810 + 100 x + 0.1 x ^{2}}{x} : 118

Minimum (90, 118)

Maximum (- 90, 82), Minimum (90, 118)

y = \frac{x ^{2} - 3}{x - 2}

6 + 2 x - 2 x^{2}

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

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

x^{2} - 2 x - 15

What is the extreme (810+100x+0.1x^2)/x ?
The extreme (810+100x+0.1x^2)/x is Maximum(-90,82),Minimum(90,118)