# Calculus I Problem Involving Tangents

1. Jan 29, 2006

### erok81

Ok, I cannot figure out how to this problem. It is the last on my practice test and have had no real problems with any of these other ones.

I have been using this formula:

$$\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}$$

Here is the problem, find the equation of the line tangent to the function at the given point.

$$f(x)=2x^3-3$$ @ (3,51)

$$y=54x-147$$ or $$y=12x-21$$

I have no clue how he gets those. Well, actually I can't even solve it anyway so that doesn't matter.

I think there might be some easier way using derivatives, but we can't use those yet since we haven't learned them. The only other formula I know of is similar to the one above but involves "h" and I can't figure out how to even use that one.

Someone show me how to this one. It might be because I forgot how to factor cubed stuff but who knows, I am so lost.

Thanks for the help.

Last edited: Jan 29, 2006
2. Jan 29, 2006

### TD

It would indeed be a lot easier if you could just use the standard rules rather than the definition. Is the problem finding the derivative or only setting up the equation for the tangent line?

3. Jan 29, 2006

### sporkstorms

This limit: $$\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}$$ gives you the slope of the line tangent to f(x) at f(x=a).

Once you know the slope, use what you learned in algebra class to find the equation of a line with that slope passing through the given point (hint, y=mx+b ;)

Is that a typo, and supposed to be the point (3, 15) ?

4. Jan 29, 2006

### TD

51 is correct since 2*3³-3 = 2*27-3 = 54-3 = 51.

5. Jan 29, 2006

### erok81

I haven't had any trouble with any of the other problems. I can find the slope and then the equation on all but this one.

That would be nice if it's a typo, then I can see why I can't get his answers. But on the practice test, the points are (3,51) I double checked just to make sure.

6. Jan 29, 2006

### erok81

I can plug it into the formula I have to use, but after that I am totally stuck. For some reason I can't figure out what to do after that to solve it. Which sucks, because these are so easy to do, yet I have spent sooo much time on this one problem.

7. Jan 29, 2006

### TD

The formula for the derivative of for the tangent line?

8. Jan 29, 2006

### sporkstorms

Yea, sorry. It's been a long night (and into today).
I just didn't like it because his tangent lines don't pass through that point.

9. Jan 29, 2006

### TD

That's right, didn't even check that yet. The first one has the correct slope though...

10. Jan 29, 2006

### erok81

This may sound mighty stupid, but I can't get past after pluggin in all the values I have.

This is as far as I can get:

$$\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}$$

$$\lim_{x\rightarrow 3} \frac{2x^3-3-51}{x-3}$$

$$\lim_{x\rightarrow 3} \frac{2x^3-54}{x-3}$$

I think the whole problem here is I don't know what to do because of the $$x^3$$ part. If I could get the slope, I can get the rest. I am just stuck getting the slope part.

11. Jan 29, 2006

### TD

Try factorising the nominator, who knows whether a factor (x-3) might pop up

12. Jan 29, 2006

### shmoe

$$x-3$$ divides $$2x^3-54$$, time for some polynomial division.

13. Jan 29, 2006

### erok81

I can't remember how to unfactor a cube anymore. I can factor one but can't go backwards anymore. :rofl:

I figured that might be part of it since there was a x-3 on the bottom. So rather than doing it the correct way, I have taken every possible combination of x-3's and x+3's and can never get back to the original equation.

Maybe it's time to get out my old algebra book and relearn factoring.:tongue2:

14. Jan 29, 2006

### TD

Well 2x³-54 = 2(x³-27) = 2(x³-3³) and that last is a difference of cubes for which you have a formula.
If you find this factoring hard, you could go for shmoe's suggestion and do the polynomial division.

15. Jan 29, 2006

### erok81

Oh yeah, that is how you do those. I didn't even think of that.

So now I can get the slope of 54. But when I put it into the equation I don't get either of those two answers. Unless I am doing it wrong, which I probably am.

Here is what I am using...

$$y-y_1=m(x-x_1)$$

$$y-51=54(x-3)$$

$$y=54x-162+51$$

$$y=54x-111$$

What else am I doing wrong?

16. Jan 29, 2006

### TD

17. Jan 29, 2006

### erok81

Oh nice, that is a relief. I was getting worried that I couldn't do them.

Well, thanks for the help everyone. Now hopefully I can do this on my test tomorrow.

I'll have to try that way of factoring you did. I never learned that way and it seems a lot shorter than dividing them.

18. Jan 29, 2006

### TD

Well, in general factoring a cubic is not so easy unless you know a root, say x = a. In that case, (x-a) is a factor and you can find the remaining quadratic using Horner's rule or by working out (x-a)(bx²+cx+d) where you have to determine b,c and d.

In this case, it's even easier since - as I said - we could write the nominator as 2(x³-3³). You can then use the formula (x³-a³) = (x-a)(x²+ax+a²) and there you immediately see the factor (x-a) which is here (x-3).

19. Jan 29, 2006

### erok81

Nice.

Thanks for explaining that. I'll give it a try on the next problem I come across.

20. Jan 29, 2006

### TD

You're welcome, good luck