What is the extreme (x^2-12)/(x+4) ?

The extreme (x^2-12)/(x+4) is Maximum(-6,-12),Minimum(-2,-4)

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extreme (x^2-12)/(x+4)

extreme points

\frac{x ^{2} - 12}{x + 4}

Maximum (- 6, - 12), Minimum (- 2, - 4)

Solution steps

f^{'} (x) = \frac{x ^{2} + 8 x + 12}{( x + 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < - 4,

Decreasing

: - 4 < x < - 2,

Increasing

: - 2 < x < \infty

Plug

x = - 6

into

\frac{x ^{2} - 12}{x + 4} : - 12

Maximum (- 6, - 12)

Plug

x = - 2

into

\frac{x ^{2} - 12}{x + 4} : - 4

Minimum (- 2, - 4)

Maximum (- 6, - 12), Minimum (- 2, - 4)

f (x) = x^{4} - 8 x^{3} + 16 x^{2}

f (x) = 16 + 4 x - x^{2}, 0 \leq x \leq 5

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

f (x, y) = x^{4} + y^{4} - 16 x y

What is the extreme (x^2-12)/(x+4) ?
The extreme (x^2-12)/(x+4) is Maximum(-6,-12),Minimum(-2,-4)