# Suppose that f(x)= (3)/((x^2)-9) A) Find all critical values of f, compute their average, (B) Use interval notation to indicate where f(x) is increasing. (C) Use interval notation to indicate where f(x) is decreasing. (D) Find the x-coordinates of all local maxima of f, compute their average. (E) Find the -coordinates of all local minima of f, compute their average. (F) Use interval notation to indicate where f(x) is concave up. (G) Use interval notation to indicate where f(x) is concave down. (H) Find the x-coordinates for all inflection points of f, compute their average. (I) Find all horizontal asymptotes of f, compute the average of the y values. (J) Find all vertical asymptotes of f, compute the average of the x values.

`f(x)=3/(x^2-9)=>`

`f'(x)=[0*(x^2-9)-3*(2x)]/(x^2-9)^2=>`

`f'(x)=(-6x)/(x^2-9)^2`

Hence the only critical point is x=0

B) Increasing

`f'(x)>0=>-6x>0=>x<0` Hence the function increase over `(-oo,0)`

C) Decreasing

f(x) decreases over `(0,oo)`

D & E) We only have one critical point x=0. f(x) increases on the left of zero and decreses on the right of zero. Hence ...

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`f(x)=3/(x^2-9)=>`

`f'(x)=[0*(x^2-9)-3*(2x)]/(x^2-9)^2=>`

`f'(x)=(-6x)/(x^2-9)^2`

Hence the only critical point is x=0

B) Increasing

`f'(x)>0=>-6x>0=>x<0` Hence the function increase over `(-oo,0)`

C) Decreasing

f(x) decreases over `(0,oo)`

D & E) We only have one critical point x=0. f(x) increases on the left of zero and decreses on the right of zero. Hence x=0 is a local max.

`f''(x)=[-6(x^2-9)^2-(-6x)*2*(x^2-9)*2x]/(x^2-9)^4=>`

`f''(x)=[(x^2-9)(-6(x^2-9)+24x^2)]/(x^2-9)^4=>`

`f''(x)=(18x^2+54)/(x^2-9)^3`

Numerator of f''(x) is always positive.

F) f(x) concave up when f''(x)>0=>`x^2-9>0`

=>(x-3)(x+3)>0=>either x-3>0 and x+3>0 or x-3<0 and x+3<0=>

x>3 and x>-3 or x<3 and x<-3 =>x>3 0r x<-3.

Hence concave up on `(-oo,-3)U(3,oo)`

G) f(x) concave down on (-3,3)

H) No point of inflection

I) Horizontal Asymptotes

None

J) Vertical Asymptotes

x=3 and x=-3

The graph below confirms our findings.

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