Results 1 to 2 of 2

Math Help - differntiable functions

  1. #1
    Member
    Joined
    Oct 2007
    From
    Nevada
    Posts
    188

    differntiable functions

    The twice-differentiable function f is defined for all real numbers and satisies the following conditions:

    f(0) = 2
    f'(0) = -4
    f''(0) = 3

    The function g is given by g(x) = e^(ax) + f(x) for all real numbers, where a is a constant. Find g'(0) and g''(0) in terms of a.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by DINOCALC09 View Post
    a) The twice-differentiable function f is defined for all real numbers and satisies the following conditions:

    f(0) = 2
    f'(0) = -4
    f''(0) = 3

    b) The function g is given by g(x) = e^(ax) + f(x) for all real numbers, where a is a constant. Find g'(0) and g''(0) in terms of a.
    Hello,

    to a):

    Assume that f''(x) = 3 then

    f'(x) = \int(f''(x))dx = 3x+C. With f'(0) = -4 you'll get C = -4

    Thus f'(x) = 3x-4

    Then f(x) = \int(f'(x)dx = \frac32 x^2-4x+C . With f(0) = 2 you'll get C = 2 and therefore the function is:

    f(x) = \frac32 x^2 -4x + 2

    to b)
    I assume that you have to use the function f from a). Thus

    g(x) = e^{ax}+f(x)~\implies~g'(x)=a \cdot e^{ax}+f'(x)~\implies~g''(x)=a^2 \cdot e^{ax}+f''(x) Plug in x = 0 (Remember e^0 = 1) and the values for f'(0) and f''(0).

    g'(0) = a \cdot 1 -4 = a-4

    g''(x) = a^2+3
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: February 23rd 2010, 04:54 PM
  2. Replies: 11
    Last Post: November 15th 2009, 11:22 AM
  3. Replies: 7
    Last Post: August 12th 2009, 04:41 PM
  4. Differntiable problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 10th 2008, 11:01 AM
  5. more differntiable
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 22nd 2007, 01:46 PM

Search Tags


/mathhelpforum @mathhelpforum