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

  • #1
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!
 

Answers and Replies

  • #2
Dick
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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.
 
  • #3
But isn't the f(a) part 0?
 
  • #4
Ray Vickson
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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).
 
Last edited:
  • #5
So can I plug in 0 for a? and then factor out h, and then plug in 0 for h?
 
  • #6
Ray Vickson
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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.
 
  • #7
But you can't because then you're going to get 0/0, right?
 
  • #8
CAF123
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What do a and h represent in your equation?
 
  • #9
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.
 
  • #10
Dick
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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.
 
  • #11
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?
 
  • #12
Dick
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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.
 
  • #13
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?
 
  • #14
Dick
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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.
 
  • #15
Ray Vickson
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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.
 

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