Differentiable function

1. Oct 2, 2008

Johny 5

1. The problem statement, all variables and given/known data
Let f be a function differentiable function with f(2) = 3 and f'(2) = -5, and let g be the function defined by g(x) = xf(x). Which of the following is an equation of the line tangent to the graph of g at point where x=2?
a. y=3x
b y-3 = -5(x-2)
c y-6 = -5(x-2)
d y-3 = -7(x-2)
e y-6 = -10(x-2)

2. Relevant equations

3. The attempt at a solution
given:
x=2
y=3
so
3=-5(2)+b
b=13
so
g(x) = x(-5x+13)
g(x) = -5x^2 + 13x
g'(x) = -10x +13
g'(2) = -7
point slope form:
d. y-3 = -7(x-2)
is that correct?

2. Oct 2, 2008

Rake-MC

What you've done here is taken the tangent line to f(x) at point x = 2 and assumed it to be f(x).

3. Oct 2, 2008

Johny 5

so the whole thing is incorrect?

4. Oct 2, 2008

sutupidmath

$$g'(x)=f(x)+xf'(x)$$

$$g'(2)=f(2)+2f'(2)=3-10=-7$$ this will be the slope of the tangent line at x=2, on the graph of g. now slope-intercept formula for a line is

$$g-g(x_0)=k(x-x_0)=>g-g(2)=-7(x-2)$$ and youre almost done.

Just find what g(2) is, substitute it, and bummmm, that's your answer.

5. Oct 2, 2008

Rake-MC

Well you tell me. Firstly you did not finish solving your equation, you stopped as soon as you saw that g'(2) = -7 and assumed that was the answer. If you had solved it all you would see that you get y - 6 = -7(x - 2). This is not an option.

I will however give you a starter:

y = -5x + 13 is NOT the function f(x), it is the equation of a tangent and only applies as far as we know at point x = 2.
Edit: Scratch that - stupid error.

Last edited: Oct 2, 2008
6. Oct 2, 2008

Johny 5

g(2) = (2)f(2)
g(2) = 2*3
g(2) = 6... hmm thats not an option

7. Oct 2, 2008

Johny 5

I stopped at g'(2) = -7 because that would be the slope, m, (you got it too) and i just plugged the given into the point slope formula (x = 2 and y = 3)

8. Oct 2, 2008

sutupidmath

Hmmm.... i just glanced at the options now..... i don't know, as far as i am concerned, if the problem is exactly phrased like this, then i guess they should put this one as an option too. Are u sure that you have written everything correctly?

9. Oct 2, 2008

sutupidmath

Your y coordinate won't be 3, since you are trying to find the slope of the tangent line at x=2 on the graph of g, so your point actually will be (2,g(2)) which is (2, 6). At least this is what you wrote on the problem on the first post.

10. Oct 2, 2008

Johny 5

yup the problem is word for word from my worksheet..... (i just re-read)

11. Oct 2, 2008

Rake-MC

However, this is the formula for f(x) not g(x) no?

12. Oct 2, 2008

sutupidmath

HOnestly, it might also be since it is 4 in the morning here, that i am missing sth really obvious there, or i really think that there is some kind of missinterpretation of that problem. Since there is no other way of goin about it, as far as i am concerned. so just add another
option

f)y-6=-7(x-2)

13. Oct 2, 2008

Rake-MC

But that would make three of us reading it incorrectly, and it's not 4am here..

14. Oct 2, 2008

Johny 5

it's for xf(x) (which is g(x)) or at least i thought it was... but you pointed out what i wrote as a function of f(x) is not f(x) its just for f(x) at x=2

15. Oct 2, 2008

sutupidmath

Well, with this i just wanted to leave a 0.01% chance that i am getting the whole thing wrong. cuz i am quite sure that the way i did it is the correct one, cuz it seems a quite straightforward problem.

16. Oct 2, 2008

sutupidmath

So, are u sayin' now that you want to find the tangent line of f(x) at x=2, or what?

if so then it would be b) y-3 = -5(x-2)

17. Oct 2, 2008

Rake-MC

That's what I got too, but I do believe that they want the tangent of g(x) at x=2. For which we can't get a answer that complies with the options

18. Oct 2, 2008

sutupidmath

We cannot get an answer that complies withthe options because there is no such answer!

19. Oct 2, 2008

Johny 5

so the answer is y-6 = -7(x-2)?... which isn't an option....

20. Oct 2, 2008

Rake-MC

as far as me and sutupidmath can see, yes.