What is the extreme f(x)=((x-5)^2)/(x+9) ?

The extreme f(x)=((x-5)^2)/(x+9) is Maximum(-23,-56),Minimum(5,0)

Popular >

extreme f(x)=((x-5)^2)/(x+9)

extreme points

f (x) = \frac{( x - 5 ) ^{2}}{x + 9}

Maximum (- 23, - 56), Minimum (5, 0)

Solution steps

f^{'} (x) = \frac{x ^{2} + 18 x - 115}{( x + 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 23,

Decreasing

: - 23 < x < - 9,

Decreasing

: - 9 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 23

into

\frac{( x - 5 ) ^{2}}{x + 9} : - 56

Maximum (- 23, - 56)

Plug

x = 5

into

\frac{( x - 5 ) ^{2}}{x + 9} : 0

Minimum (5, 0)

Maximum (- 23, - 56), Minimum (5, 0)

f (x) = x^{5} - 2 x^{3} + 1

x^{4} - 8 x^{3} + 6

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

f (x) = - x + cos (3 π x)

f (x) = 3 x^{3} - 36 x^{2} + 108 x + 8, 0 \leq x \leq 9

What is the extreme f(x)=((x-5)^2)/(x+9) ?
The extreme f(x)=((x-5)^2)/(x+9) is Maximum(-23,-56),Minimum(5,0)