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Homework Help: Differentiable function

  1. Oct 2, 2008 #1
    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
    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. jcsd
  3. Oct 2, 2008 #2
    What you've done here is taken the tangent line to f(x) at point x = 2 and assumed it to be f(x).
  4. Oct 2, 2008 #3
    so the whole thing is incorrect?
  5. Oct 2, 2008 #4

    [tex]g'(2)=f(2)+2f'(2)=3-10=-7[/tex] 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

    [tex]g-g(x_0)=k(x-x_0)=>g-g(2)=-7(x-2)[/tex] and youre almost done.

    Just find what g(2) is, substitute it, and bummmm, that's your answer.
  6. Oct 2, 2008 #5
    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.
    Read what sutupidmath said.
    Last edited: Oct 2, 2008
  7. Oct 2, 2008 #6
    g(2) = (2)f(2)
    g(2) = 2*3
    g(2) = 6... hmm thats not an option
  8. Oct 2, 2008 #7
    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)
  9. Oct 2, 2008 #8
    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?
  10. Oct 2, 2008 #9
    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.
  11. Oct 2, 2008 #10
    yup the problem is word for word from my worksheet..... (i just re-read)
  12. Oct 2, 2008 #11
    However, this is the formula for f(x) not g(x) no?
  13. Oct 2, 2008 #12
    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

  14. Oct 2, 2008 #13
    But that would make three of us reading it incorrectly, and it's not 4am here..
  15. Oct 2, 2008 #14
    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
  16. Oct 2, 2008 #15
    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.
  17. Oct 2, 2008 #16
    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)
  18. Oct 2, 2008 #17
    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
  19. Oct 2, 2008 #18

    We cannot get an answer that complies withthe options because there is no such answer!
  20. Oct 2, 2008 #19
    so the answer is y-6 = -7(x-2)?... which isn't an option....
  21. Oct 2, 2008 #20
    as far as me and sutupidmath can see, yes.
  22. Oct 2, 2008 #21
    i just graphed the function that we got to be g(x), which is xf(x)
    C.y-6 = -5(x-2) was a tangent to the line y=x(-5x+13) at x=2 (2,6)

    NOTE: y-6 = -7(x-2) also got the same results at x=2??!? this is the equation that wasn't an option... the other one is but i don't know how to get there
  23. Oct 2, 2008 #22
    First i don't know how could you graph g(x) when we don't actually know at first place what f(x) is, we could try to figure that out, but still.
    I am completely confused: A tangent to the line :y=x(-5x+13) , where did u get this one from? Or, just don't tell me that your f(x) =-5x+13, and u didn't tell us this from the very begginning?
  24. Oct 2, 2008 #23
    if thats the case then, [tex]g(x)=-5x^2+13x[/tex] so [tex]g'(x)=-10x+13[/tex] so [tex]g'(2)=-7[/tex]

    [tex]g(2)=6[/tex] which would give us again exactly the same thing. since

  25. Oct 2, 2008 #24
    What he has written is incorrect. y=-5x+13 is the equation for the tangent of f(x) ONLY at point x=2 (as far as we know).

    [tex] f(2) = 3 [/tex]

    [tex] f'(2) = -5 [/tex]

    [tex] y = mx + c [/tex]

    [tex] 3 = -5(2) + c [/tex]

    [tex] c = 13 [/tex]

    tangent for f(x) at point x=2 is: [tex] y = -5x +13 [/tex]

    I believe he has become confused and thought this is the equation for the function f(x) and worked from there.
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