Calculus I Problem Involving Tangents

[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:

[tex]\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}[/tex]


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

[tex]f(x)=2x^3-3[/tex] @ (3,51)

The answers he shows are..

[tex]y=54x-147[/tex] or [tex]y=12x-21[/tex]

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.

 

[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?

 

[sporkstorms]

This limit: [tex]\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}[/tex] 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 ;)

[tex]f(x)=2x^3-3[/tex] @ (3,51)
The answers he shows are..
[tex]y=54x-147[/tex] or [tex]y=12x-21[/tex]

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

 

[TD]

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

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

 

[erok81]

This limit: [tex]\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}[/tex] 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) ?

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.

 

[erok81]

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?

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.

 

[TD]

The formula for the derivative of for the tangent line?
Please show us where you get stuck, then I'll help you further.

 

[sporkstorms]

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

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.

 

[TD]

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.

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

 

[erok81]

The formula for the derivative of for the tangent line?
Please show us where you get stuck, then I'll help you further.

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:

[tex]\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a}[/tex]

[tex]\lim_{x\rightarrow 3} \frac{2x^3-3-51}{x-3}[/tex]

[tex]\lim_{x\rightarrow 3} \frac{2x^3-54}{x-3}[/tex]

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

 

[TD]

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

 

[shmoe]

[tex]\lim_{x\rightarrow 3} \frac{2x^3-54}{x-3}[/tex]

[tex]x-3[/tex] divides [tex]2x^3-54[/tex], time for some polynomial division.

 

[erok81]

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

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

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.

 

[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.

 

[erok81]

[tex]x-3[/tex] divides [tex]2x^3-54[/tex], time for some polynomial division.

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...

[tex]y-y_1=m(x-x_1)[/tex]

[tex]y-51=54(x-3)[/tex]

[tex]y=54x-162+51[/tex]

[tex]y=54x-111[/tex]

What else am I doing wrong?

 

[TD]

Your answer is correct, the given answers were wrong (as pointed out by sporkstorms earlier).

 

[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.

 

[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).

 

[erok81]

Nice.

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

 

[TD]

You're welcome, good luck

 

[HallsofIvy]

Going back to your first post- that's the reason sporkstorms asked if that wasn't (3, 15). When x= 3, both y=54x-147 and y= 12x- 5 give y= 15, not 51.

But, as TD pointed out, when x= 3 in the original equation,
[tex]f(x)=2x^3-3[/tex]
y= 51.

I don't know who "he" is or why he would list two different answers for a problem with one answer but neither of them is correct- your equation is correct.

 

[erok81]

Going back to your first post- that's the reason sporkstorms asked if that wasn't (3, 15). When x= 3, both y=54x-147 and y= 12x- 5 give y= 15, not 51.

But, as TD pointed out, when x= 3 in the original equation,
[tex]f(x)=2x^3-3[/tex]
y= 51.

I don't know who "he" is or why he would list two different answers for a problem with one answer but neither of them is correct- your equation is correct.

He is my math teacher. Not the best I have had, but better than nothing I guess. This was all of the pre-test the teacher made for us to study with.

No wonder I could never get the answer right since the answer on the test was incorrect to start with. Oh well, at least I know what it should be now.