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Problme with word problems

  1. Jul 18, 2006 #1
    If I am asked to find the constants a, b, c, d such that the graph of
    f(x)= ax^3 + bx^2 + cx+ d has horizontal tangent lines at the points (-2, 1) and (0, -3).

    I am not sure what to go about doing it...is that asking me to find what? (I dont know what do the constant stand for? in form of y= mx+b?)

    I know though, the first thing I would do is, find the derivative of the function..."f(x)= ax^3 + bx^2 + cx+ d "

    which would be...

    f' (x) = 3 ax^2+ 2 bx+ c
    then sub the value of x? x = (-2) into the last equation..
    which will equal to
    12 a- 4b+ c

    nowwwwww? what do I do next?:grumpy:

    and I have another q.

    Given h= f 0 g, g(3)=7, g'(3)=4, f(2)=4, f'(7)=-6.
    now how do I determine the h' (3)???

    again i am half way through the answer...

    I think thinking of solving it with product rule??
    h(x)= f(g) x)) h(x)= f' (g(x) g'(x)

    h' (x)= f' (g (3)= g'3= f' (7) (4) = (-6) (-4) = -24

    soooo plz help???:confused:
  2. jcsd
  3. Jul 18, 2006 #2


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    You've done the first problem correctly so far. You know the tangent line is horizontal when the derivative is zero, so you know the equation you found:

    f' (x) = 3 ax^2+ 2 bx+ c

    must be zero at x=-2. You also know the original equation,

    f(x)= ax^3 + bx^2 + cx+ d

    must be 1 at x=-2, as given.

    You also have another set of equations, with f'(x) = 0 and f(x) = -3 at x=0.

    Solve the simultaneous equations.

    - Warren
  4. Jul 18, 2006 #3
    Hi Warren, thanks for replying..
    But I dont get the part where you think _ must be zero ( do u mean the x value ?)

    and how do u figure the value of 1 @ x= -2?

    i am lost..I m taking a summer course it is going at fast as speed of the light..it is hard to absorb the material.:surprised
  5. Jul 18, 2006 #4


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    If there is a horizontal tangent at a point P=(x,y), then f '(x)=0 (the derivative)
  6. Jul 18, 2006 #5


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    Take it a bit slower... you seem to be missing some vital parentheses h'(x) = f '(g(x))*g'(x). So for x=3:

    f '(g(3))*g'(3)

    Now try solving that
  7. Jul 18, 2006 #6
    I feel like a loser..:uhh: oh well, let me give my best shot:shy:

    Since h (x)= f (g (x))
    then can I apply this ruling to solve for h (x)

    f '(g(3))*g'(3) = 9? (by multiplying?)
  8. Jul 18, 2006 #7
    Am I right?? anyone? Please
  9. Jul 18, 2006 #8


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    f '(g(3)) = f '(7) from the info. f '(7) = -6. Multiply that by g'(3), and you get -24

    You had it right the first time, I just couldn't tell because of the missing parentheses if you used the right method :redface:
  10. Jul 18, 2006 #9


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    Uhmm, ok, for the first problem:
    You have 4 unknowns, namely a, b, c, and d. So you must need 4 equation to solve for them.
    Horizontal tangent line has a slope of 0, right? Since it's parallel to the Ox axis.
    So, at x = 0, and x = -2, the graph of f(x) has horizontal tangent lines. Or in other words:
    f'(0) = f'(-2) = 0. Right?
    This is the first 2 equations, right?
    Given that the 2 points (-2, 1), and (0, -3) are on the curve. So you'll have another 2 equations:
    f(-2) = 1, and f(0) = -3.
    Having 4 equations, can you solve for the 4 unknowns?
    Can you get this? Is there anything unclear? :)
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