What is the extreme (x^2-7)/(x+4) ?

The extreme (x^2-7)/(x+4) is Maximum(-7,-14),Minimum(-1,-2)

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extreme (x^2-7)/(x+4)

extreme points

\frac{x ^{2} - 7}{x + 4}

Maximum (- 7, - 14), Minimum (- 1, - 2)

Solution steps

f^{'} (x) = \frac{x ^{2} + 8 x + 7}{( x + 4 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 7,

Decreasing

: - 7 < x < - 4,

Decreasing

: - 4 < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 7

into

\frac{x ^{2} - 7}{x + 4} : - 14

Maximum (- 7, - 14)

Plug

x = - 1

into

\frac{x ^{2} - 7}{x + 4} : - 2

Minimum (- 1, - 2)

Maximum (- 7, - 14), Minimum (- 1, - 2)

f (x) = \frac{x ^{3}}{3} + 4 x^{2} + 15 x + 7

T (x, y) = \frac{x ^{3} + y ^{3}}{( x + y ) ^{2} - 3 x y} - x

f (x) = 6 x - 6

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

What is the extreme (x^2-7)/(x+4) ?
The extreme (x^2-7)/(x+4) is Maximum(-7,-14),Minimum(-1,-2)