What is the extreme f(x)=(x^2+69)/(x+10) ?

The extreme f(x)=(x^2+69)/(x+10) is Maximum(-23,-46),Minimum(3,6)

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extreme f(x)=(x^2+69)/(x+10)

extreme points

f (x) = \frac{x ^{2} + 69}{x + 10}

Maximum (- 23, - 46), Minimum (3, 6)

Solution steps

f^{'} (x) = \frac{x ^{2} + 20 x - 69}{( x + 10 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 23,

Decreasing

: - 23 < x < - 10,

Decreasing

: - 10 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 23

into

\frac{x ^{2} + 69}{x + 10} : - 46

Maximum (- 23, - 46)

Plug

x = 3

into

\frac{x ^{2} + 69}{x + 10} : 6

Minimum (3, 6)

Maximum (- 23, - 46), Minimum (3, 6)

\frac{t ^{2}}{t + 3}

f (x) = x^{3} - 10 x + 25 x - 2

f (x) = \frac{20}{x} + 2 π x^{2}, x > 0

f (x) = - x^{3} + x^{2} + 11 x - 1

f (x) = 2 x - sin (x)

What is the extreme f(x)=(x^2+69)/(x+10) ?
The extreme f(x)=(x^2+69)/(x+10) is Maximum(-23,-46),Minimum(3,6)