# Suppose that f(x)= (7x-5)/(x+6)(A) Find all critical values of f. If there are no critical values, enter -1000. (B) Use interval notation to indicate where f(x) is increasing. (C) Use interval...

Suppose that f(x)= (7x-5)/(x+6)

(A) Find all critical values of f. If there are no critical values, enter -1000.

(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. If there are no local maxima, enter -1000.

(E) Find the x-coordinates of all local minima of f. If there are no local minima, enter -1000.

(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 all inflection points of f. If there are no inflection points, enter -1000.

(I) Find all horizontal asymptotes of f. If there are no horizontal asymptotes, enter -1000.

(J) Find all vertical asymptotes of f. If there are no vertical asymptotes, enter -1000.

### 1 Answer | Add Yours

`f(x)=(7x-5)/(x+6)`

Let's start with I and J

I) Horizontal Asymptote

`lim_(x->oo)f(x)=7/1=7`

Hence y=7 is the horizontal asymptote

J) Vertical Asymptote

`lim_(x->-6)f(x)=oo`

Hence x=-6 vertical asymptote.

`f'(x)=[7(x+6)-(7x-5)*1]/(x+6)^2=>`

`f'(x)=[7x+42-7x+5]/(x+6)^2=47/(x+6)^2`

A) Since f(x) is not defined at x=-6, then we have no critical points.

B & C) f'(x) is always positive, hence it is always increasing.

`f''(x)=[-47*2*(x+6)]/(x+6)^4=-94/(x+6)^3`

D & E) no local max or min

F) f(x) is concave up when f''(x)>0=>`(x+6)<0=>x<-6`

`(-oo,-6)`

G) f(x) is concave down on `(-6,oo)`

H) No inflection point.