10 points to who can help me with derivatives and other nonsense?

Question

Let f(x) = 3/(x - 1)

a. find the slope of the tangent to the graph of f at a general point xsub0.

b. use the result of part a. to find the slope of the tangent at xsub0 = 4

c. write the equation of the line tangent to the curve at xsub0 = 4

see the problem is that i cant even get the first part.. so i cant get the second or third.

whats really confusing me is that after you plug it into the:

lim as x -> 0 (f(x1 + h) - f(x1))/h

i dont know how to simplify it.

thanks in advance!

Answer 1

a. first make it 3(x-1)^-1 then do chain rule 3u^-1=-3u^-2du (du=1)

dy/dx=-3/(x-1)^2

b. -3/(4-1)^2

-3/9

-1/3

c. you have a slope, now you need a point. plug in 4 to f(x) to get point.

3/(4-1)=1 so you have point (4,3)

then plug in to y=mx+b and solve for b

3=(-1/3)(4)+b

b=13/3

y=-1/3x+13/3

make it a good day

Answer 2

f(x) = 3/(x-1) = 3(x-1)^-1

f'(x) = -3(x-1)^-2 = -3/(x-1)^2  (From chain rule)

a) At xsub0, the gradient of the tangent is -3/(xsub0-1)^2

b) When xsub0 = 4, f'(x) = -1/3

c) y = f(x), known point on line = (4,1)

Gradient = -1/3

-1/3 = (y-1)/(x-4)

y = 7/3 - x/3

Just use the chain rule with u=x-1 to do the first part.

EDIT: climberg made an error in his known point, it is (4,1) not (4,3)