`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.