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Cal Test tom, few questions

  1. Jul 9, 2006 #1
    Im cramming for my cal test tomorrow, and im having a brain fart right now.

    Heres a few problems on my Cal review i need some help with:

    If f(5)=3 and f'(5)=2, use differentials to approx. f(5.01) - no clue what to do, i just need a point in the right direction

    Let f(x) = x^3+2x^2-1 Suppose x = 2 and dx = 0.02

    A) Calculate delta y
    B) Calculate dy - already done

    I solved part b using the equation dy=f'(x)dx but im stuck on part a
    Last edited: Jul 9, 2006
  2. jcsd
  3. Jul 9, 2006 #2
    no.10 : use the definition of the derivative and use delta x = 0.01

    no.7 : a ) Here again, use the definition of the derivative
    b ) great !
    Hope it helps
  4. Jul 9, 2006 #3

    For the second question I think it's just asking for the difference between the y values at those two points in other words the numerical difference between f(x + dx) and f(x).

    For the first question you should know that you can approximate a function around any point where you know the value of the function at that point and the slope of the tangent line at that point, what you're doing is using the tangent line to the function as an approximation to the actual function.

    The derivative is defined as the limit of (f(x + h) - f(x))/h as h goes to zero. Now for small h but nonzero h we should be able to say that

    f'(x) is approximately equal to (f(x+h) - f(x))/h

    and since you have f(5) and f'(5) you can approximate f(5.01).
  5. Jul 9, 2006 #4
    i got question #7

    for question #10 however,

    f(5)=3 f'(5)=2 approx. f(5.01)=?

    im still confused how do my two equations help me solve the problem?

    delta y = f(x+delta x) - f(x)
    dy = f'(x)dx
    Last edited: Jul 9, 2006
  6. Jul 9, 2006 #5

    Ok using those equations you have that f(x + delta x) - f(x) = delta y

    and dy = f'(x)dx

    But for relatively small x delta y and dy are approximately equal so we can substitute the second equation into the first giving

    f'(x)dx = f(x + delta x ) - f(x)

    so you approximate f(x + delta x ) by the equations above as
    f(x + delta x ) = f(x) + f'(x)*(delta x) because we are assuing that for small x dx is approximately equal to delta x.
  7. Jul 9, 2006 #6
    Thanks for all the help, almost finished:

    14. Find the function f that satisfies the following conditions:

    f ''(x) = sin x , f'(0)=2 , f(pi) = -1

    I understand u can take the anti deravitive of f ''(x) and then solve for C then, take the anti deravitive of that ( f(x) ). But im stuck because there isnt an anti deravitive of sin x. Am i setting this problem up wrong?
  8. Jul 9, 2006 #7
    What is the derivitave of cos(x)? How about the derivative of -cos(x)?
  9. Jul 9, 2006 #8
    so -f'(x) = cos x ? therefore f'(x) = -cos x?
  10. Jul 9, 2006 #9
    Well since you started with f"(x) = sin(x) you technically have f'(x) = -cos(x) +C, and since you know f'(0) =2 you can solve for C then integrate again and repeat the same process to find f(x).
  11. Jul 9, 2006 #10


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    Why do you say there is no antiderivative of sinx? It's simply -cos(x) +C !
    Then you can integrate this again.

    EDIT: apologies to d leet! Had not seen your post!
  12. Jul 9, 2006 #11

    ahh that makes sense, thanks alot. both of you thanks
  13. Jul 9, 2006 #12
    Glad to help, and it's good to hear that it makes sense now. Good luck on your test tomorrow.
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