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Math Help - How do I start this derivative problem?

  1. #1
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    How do I start this derivative problem?

    The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

    The graph:
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  2. #2
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    Re: How do I start this derivative problem?

    Find each of the following what? You already showed the graph, so you should be able to figure out f(x) and g(x) as piecewise continuous functions.

    f(x) = \left\{\begin{array}{ll}m_1 x + b_1 & -1<x<2 \\ m_2 x + b_2 & 2<x<6 \end{array} \right.

    g(x) = \left\{\begin{array}{ll}m_3 x + b_3 & -1 < x < 0 \\ m_4 x + b_4 & 0 < x < 3 \\ m_5 x + b_5 & 3 < x < 6 \end{array} \right.

    Find the five slopes and five corresponding y-intercepts (to find the y-intercepts, you pretend that each line segment continues to figure out where it would hit the y-axis if it were a line). To be the most accurate, try to find specific points that the line segments cross. For example, the left-most red line segment passes between the points (-1,5) and (1,2), so you can use those points to find the slope of that line segment, which would tell you m_1.

    What else do you need help figuring out?
    Thanks from topsquark
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  3. #3
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    Re: How do I start this derivative problem?

    So how would I find u'(x) from u(x)? and v'(x) from v(x)?
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    Re: How do I start this derivative problem?

    Quote Originally Posted by rhcprule3 View Post
    So how would I find u'(x) from u(x)? and v'(x) from v(x)?
    There are at least two ways of finding those derivatives. One way: use the product rule to find u'(x) and the quotient rule to find v'(x). Another way: write out the piecewise functions and differentiate. For example:

    u(x) = \left\{ \begin{array}{ll} (m_1 x + b_1)(m_3 x + b_3) & -1 < x < 0 \\ (m_1 x + b_1)(m_4 x + b_4) & 0 < x < 2 \\ (m_2 x + b_2)(m_4 x + b_4) & 2 < x < 3 \\ (m_2 x + b_2)(m_5 x + b_5) & 3 < x < 6\end{array} \right.

    And similarly for v(x).
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  5. #5
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    Re: How do I start this derivative problem?

    The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

    u(1) =

    v(1) =

    this is the actual question

    so to find u'(1), I would need to use the product rule?
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  6. #6
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    Re: How do I start this derivative problem?

    Quote Originally Posted by rhcprule3 View Post
    The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

    u(1) =

    v(1) =

    this is the actual question

    so to find u'(1), I would need to use the product rule?
    Now I am confused. Do you want u(1) or u'(1)?

    Assuming you want u'(1), do you know the product rule? If you do, then find u'(x) and plug in 1.
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  7. #7
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    Re: How do I start this derivative problem?

    Quote Originally Posted by rhcprule3 View Post
    The graphs of the function f (given in blue) and g (given in red) are plotted above. Suppose that u(x)=f(x)g(x) and v(x)=f(x)g(x). Find each of the following:

    u(1) =

    v(1) =

    this is the actual question

    so to find u'(1), I would need to use the product rule?
    v(x)=f(x)g(x)
    u(x)=f(x)g(x)

    What is the distinction between u and v? Are you sure this is product problem or it is a composition problem, that is
    u(x) = f(x) o g(x)
    v(x) = g(x) o f(x)
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  8. #8
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    Re: How do I start this derivative problem?

    If I read it correctly (and there are some strange characters in the post), u= f times g and v= f divided by g.

    Yes, by the product rule, u'= f'g+ fg' and by the quotient rule, v'= (f'g- fg')/v^2. Now, from the graph, f' is -3 for x< 0, 2 for 0< x< 3, and -1 for x> 3 while g' is -3/2 for x< 2 and 1/2 for x> 2.
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  9. #9
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    Re: How do I start this derivative problem?

    Quote Originally Posted by HallsofIvy View Post
    If I read it correctly (and there are some strange characters in the post), u= f times g and v= f divided by g.

    Yes, by the product rule, u'= f'g+ fg' and by the quotient rule, v'= (f'g- fg')/v^2. Now, from the graph, f' is -3 for x< 0, 2 for 0< x< 3, and -1 for x> 3 while g' is -3/2 for x< 2 and 1/2 for x> 2.
    I went backward to the first post, you got me, it looks like a division OP rewrite it as product in the fifth post, probably a typo. That confused me.
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