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Thread: Calculus/Differentiability

  1. #1
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    Calculus/Differentiability

    Hello, I need help finishing this problem. Use one-sided limits to find values of the constants a and b that make the piecewise function differentiable at the point where the function rule changes.

    f(x) = e^(ax), if x <1
    and
    f(x) = b + ln x, if x>1

    Here's what I have so far.....

    For f to be continuous at x = 1,
    lim as x goes to 1 from the left of e^(ax)= lim as x goes to 1 from the right of (b+ ln x)

    implies e^(a)=b

    For f to be differentiable at x = 1,
    lim as x goes to 1 from the left of ae^(ax)=lim as x goes to 1 from the right of (1/x)

    implies ae^(a) = 1

    So I'm having trouble solving for a and b for these two equations....
    e^(a)=b
    ae^(a) = 1

    The solution says solve by grapher: a = 0.5671....and b = 1.7632....,
    but I don't know how to get that.

    Thanks
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  2. #2
    Ted
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    Quote Originally Posted by pantera View Post
    Hello, I need help finishing this problem. Use one-sided limits to find values of the constants a and b that make the piecewise function differentiable at the point where the function rule changes.

    f(x) = e^(ax), if x <1
    and
    f(x) = b + ln x, if x>1

    Here's what I have so far.....

    For f to be continuous at x = 1,
    lim as x goes to 1 from the left of e^(ax)= lim as x goes to 1 from the right of (b+ ln x)

    implies e^(a)=b

    For f to be differentiable at x = 1,
    lim as x goes to 1 from the left of ae^(ax)=lim as x goes to 1 from the right of (1/x)

    implies ae^(a) = 1

    So I'm having trouble solving for a and b for these two equations....
    e^(a)=b
    ae^(a) = 1

    The solution says solve by grapher: a = 0.5671....and b = 1.7632....,
    but I don't know how to get that.

    Thanks
    You function is not defined at 1.
    Are you sure there is no $\displaystyle \leq 1$ or $\displaystyle \geq 1$ in the function ?
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  3. #3
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    Actually it is f(x) = e^(ax), if x is less than or equal to one.
    Thanks.
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  4. #4
    MHF Contributor chisigma's Avatar
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    The problem is to find $\displaystyle a$ because for $\displaystyle b$ is $\displaystyle b= e^{a}$. The equation for $\displaystyle a$ is...

    $\displaystyle a\cdot e^{a}=1$ (1)

    ... that can be written as...

    $\displaystyle f(a)=a+\ln a=0$ (2)

    An efficient way to solve an equation like (2) is the Newton-Raphson method that computes the sequence...

    $\displaystyle a_{n+1} = a_{n} - \frac{f(a_{n})}{f^{'}(a_{n})}$ (3)

    If $\displaystyle a_{0}$ is chosen correctly [for example $\displaystyle a_{0}=\frac {1}{2}$...] the sequence (3) converges to the solution...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
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    Quote Originally Posted by pantera View Post
    Hello, I need help finishing this problem. Use one-sided limits to find values of the constants a and b that make the piecewise function differentiable at the point where the function rule changes.

    f(x) = e^(ax), if x <1
    and
    f(x) = b + ln x, if x>1

    Here's what I have so far.....

    For f to be continuous at x = 1,
    lim as x goes to 1 from the left of e^(ax)= lim as x goes to 1 from the right of (b+ ln x)

    implies e^(a)=b

    For f to be differentiable at x = 1,
    lim as x goes to 1 from the left of ae^(ax)=lim as x goes to 1 from the right of (1/x)

    implies ae^(a) = 1

    So I'm having trouble solving for a and b for these two equations....
    e^(a)=b
    ae^(a) = 1
    There is no "algebraic" way to solve that second equation. You can say that a= W(1) where W is the "Lambert W function" which is defined as the inverse to f(x)= xe^x. Other that that, it must be solved numerically which you seem to have done.

    The solution says solve by grapher: a = 0.5671....and b = 1.7632....,
    but I don't know how to get that.

    Thanks
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  6. #6
    Junior Member
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    Thank you so much!
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