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Thread: Tangent lines

  1. #1
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    Tangent lines

    I have a problem that I do not know how to start. If anyone could give me a hint on how to start it, that would be great.

    Let $\displaystyle f$ be the function given by $\displaystyle f(x) = e^{-x}$, and let $\displaystyle g$ be the function given by $\displaystyle g(x) = kx$, where $\displaystyle k$ is the nonzero constant such that the graph of $\displaystyle f$ is tangent to the graph of $\displaystyle g$.

    Find the x-coordinate of the point of tangency and the value of $\displaystyle k$.
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  2. #2
    Member arpitagarwal82's Avatar
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    By condition of tangance df/dx x dg/dx = -1

    so$\displaystyle -e^{-x} * k = -1
    $
    $\displaystyle e^{-x} = 1/k$

    $\displaystyle e^x = k$
    $\displaystyle x = \ln k$

    So g(x) is tangent to f(x) at point $\displaystyle x = \ln k$

    Now for x = \ln k
    you cordinates of both curve should be same

    so x = \ln k satisfies
    $\displaystyle e^{-x}= kx$

    solve for k
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  3. #3
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    Tangent

    Hello Cursed
    Quote Originally Posted by Cursed View Post
    I have a problem that I do not know how to start. If anyone could give me a hint on how to start it, that would be great.

    Let $\displaystyle f$ be the function given by $\displaystyle f(x) = e^{-x}$, and let $\displaystyle g$ be the function given by $\displaystyle g(x) = kx$, where $\displaystyle k$ is the nonzero constant such that the graph of $\displaystyle f$ is tangent to the graph of $\displaystyle g$.

    Find the x-coordinate of the point of tangency and the value of $\displaystyle k$.
    First, you need to find the value of $\displaystyle x$ where the two graphs meet: that is where $\displaystyle f(x) = g(x)$, or

    $\displaystyle e^{-x}= kx$ (1)

    Then if $\displaystyle g(x)$ is a tangent to $\displaystyle f(x)$ for this value of $\displaystyle x$, their gradients are equal at this point. So $\displaystyle f'(x) = g'(x)$ or:

    $\displaystyle -e^{-x} = k$

    $\displaystyle \Rightarrow e^{-x} = -k$ (2)

    Now you can use (2) to eliminate the $\displaystyle e^{-x}$ term in (1), and solve the resulting equation for $\displaystyle x$. This is the $\displaystyle x$-coordinate of the point of tangency, which you can then use to find the value of $\displaystyle k$, by substituting it back into (1).

    Can you complete it now?

    Grandad
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  4. #4
    Member arpitagarwal82's Avatar
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    Ok
    my solution was wrong.
    Proceed with Granded's solution.

    Thanks Granded
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