# Finding the slope of the tangent line using the f(a+h)-f(a)/h form?

Find the slope of the tangent line to the curve f(x)=x-x^3 at the point (1,0)

So, I went through and plugged (a+h) in everywhere I saw "x"

(a+h)-(a+h)^3

Factoring:

(a+h-a^3-3a^2h-3ah^2-h^3)

I'm really stuck on what to do next...I don't see anything you can cancel/pull out? I know you need to cancel the h in the denominator, but I can't pull an h out because of the "a"s...

Thank you!

Dick
Homework Helper
Find the slope of the tangent line to the curve f(x)=x-x^3 at the point (1,0)

So, I went through and plugged (a+h) in everywhere I saw "x"

(a+h)-(a+h)^3

Factoring:

(a+h-a^3-3a^2h-3ah^2-h^3)

I'm really stuck on what to do next...I don't see anything you can cancel/pull out? I know you need to cancel the h in the denominator, but I can't pull an h out because of the "a"s...

Thank you!

That's just the f(a+h) part. Subtract the f(a) part before you start trying to factor.

But isn't the f(a) part 0?

Ray Vickson
Homework Helper
Dearly Missed
But isn't the f(a) part 0?

Yes, in this case. But you have not used that information yet in your calculation of f(a+h).

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So can I plug in 0 for a? and then factor out h, and then plug in 0 for h?

Ray Vickson
Homework Helper
Dearly Missed
So can I plug in 0 for a? and then factor out h, and then plug in 0 for h?

Don't ask; just decide for yourself.

But you can't because then you're going to get 0/0, right?

CAF123
Gold Member
What do a and h represent in your equation?

a would represent the "lim as x->a" so it would be the "lim as x->1"...h is what you are trying to factor out.

Dick
Homework Helper
a would represent the "lim as x->a" so it would be the "lim as x->1"...h is what you are trying to factor out.

If you are saying a=1 in this problem, that would be right.

But I calculated the slope another way, and the answer is -2. When I do it this way, you'll get 0/0 though if you plug in 1 for a because you would then plug in 0 for the h's and have no number left?

Dick
Homework Helper
But I calculated the slope another way, and the answer is -2. When I do it this way, you'll get 0/0 though if you plug in 1 for a because you would then plug in 0 for the h's and have no number left?

Show your work. You should be able to cancel an h and then let h->0 and get -2 this way too.

I put the first steps on how I factored it out above

Then,

a+h-a^3-3a^2h-3ah^2-h^3

Plug in 1:
h-3h-3h^2-h^3

pull out an h, I get h(-3-3h-h^2)

Then you'll get 0/0?

Dick
Homework Helper
I put the first steps on how I factored it out above

Then,

a+h-a^3-3a^2h-3ah^2-h^3

Plug in 1:
h-3h-3h^2-h^3

pull out an h, I get h(-3-3h-h^2)

Then you'll get 0/0?

i) your expression will have an h in the denominator as well. Cancel it with the one in the numerator before taking h->0. ii) Your algebra isn't correct h-3h-3h^2-h^3 isn't equal to h(-3-3h-h^2). Fix it.

Ray Vickson
Homework Helper
Dearly Missed
I put the first steps on how I factored it out above

Then,

a+h-a^3-3a^2h-3ah^2-h^3

Plug in 1:
h-3h-3h^2-h^3

pull out an h, I get h(-3-3h-h^2)

Then you'll get 0/0?

There is something fundamental that you are not "getting". Here is how it should go.

(1) First you calculate [f(a+h)-f(a)]/h, where h is NOT zero. Simplify it as much as you can.
(2) Then (and only then), let h → 0 to see what is the limit.

You never actually set h = 0. Taking h → 0 is not the same as setting h = 0.