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

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

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

extreme points

f (x) = \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) = \frac{49}{2} x^{2} - ln (x)

f (x) = 4 x^{3} e^{- x}, - 1 \leq x \leq 4

x^{4} - 4 x^{2} + 2

(x^{2} - 4)^{7}

f (x, y) = \frac{1}{16 - 4 x ^{2} - y ^{2}}

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