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Differentiating to find one unknown function out of three?

  1. Aug 26, 2012 #1
    differentiating to find one unknown function out of three??

    1. The problem statement, all variables and given/known data
    Hi everyone

    I need some help with a question that i have solved yet i find it hard to understand.

    I have three given functions in the picture attached. All three consist of one single graph. And they give me the functions for two parts whereas i have to find out the last one which is the parable.



    2. Relevant equations

    I needed to differentiate what i already had given, since i wanted to find the slope of the functions since i also know that where one function stops and the other begins i have the same slope.

    3. The attempt at a solution

    I tried to solve it by saying that g(x)=ax^2+bx+c

    then i differentiate it and it becomes:

    g'(x)=2ax+b

    i now look at the first function f(x)=-1 is f^' (x)=0 this means that the slope for this function is 0.

    the slope for the points (-2;0) must therefore also be 0.

    i put this in the parable function

    g'(x)=2ax+b
    g'(-2)=-4a+b=0

    then i know that in the points (0;0) the slope of the parable must be the same as the one for h(x)




    i have to find h'(0) and it wont work unless i get it to be h'(0)=1=b but i dont get 1 since i say h'(0)=0,006*0^2-0,18
    h'(0)=-0,18

    the end result should be g(x) = 1/4 x + x

    what is wrong where i stop up ? can someone please tell my why i wont get it right..
     
    Last edited: Aug 26, 2012
  2. jcsd
  3. Aug 26, 2012 #2

    Mentallic

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    Re: differentiating to find one unknown function out of three??

    This is very hard to follow. Could you please post the question you're trying to answer in its entirety? And it's called a parabola, not a parable.
     
  4. Aug 26, 2012 #3

    Ray Vickson

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    Re: differentiating to find one unknown function out of three??

    I do not see any attached picture.

    I am assuming that when you say "i", you are referring to yourself rather than the square root of -1. I guess you mean "I".

    I find it impossible to follow what you are doing.

    RGV
     
    Last edited: Aug 26, 2012
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