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Thread: tangent line to inverse function at P

  1. #1
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    tangent line to inverse function at P

    f(x) = x^5+3x^3+2x-1

    Find the slope of the tangent line at point (5,1) on the graph of f^-1.

    Thank you.
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  2. #2
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    Hello, 2clients!

    $\displaystyle f(x) \:= \:x^5+3x^3+2x-1$

    Find the slope of the tangent line at point (5,1) on the graph of $\displaystyle f^{-1.}(x).$

    The inverse function is: .$\displaystyle y^5 + 3y^3 + 2y - 1 \:=\:x$

    Differentiate implicitly: .$\displaystyle (5y^4 + 9y^2 + 2)\frac{dy}{dx} \:=\:1 \quad\Rightarrow\quad \frac{dy}{dx} \:=\:\frac{1}{5y^4 + 9y^2 + 2}$

    At $\displaystyle (5,1)\!:\;\;\frac{dy}{dx} \:=\:\frac{1}{5\!\cdot\!1^4 + 9\!\cdot\!1^2 + 2} \:=\:\boxed{\frac{1}{16}}$

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  3. #3
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    Well, you have to know that:
    If f is differentiable at all x and f has an inverse function $\displaystyle g(x) = f^{-1}(x)$, then:
    $\displaystyle g'(x) = \frac{1}{f'(g(x))}$

    where $\displaystyle f`(g(x)) \neq 0$

    ===============


    We need to find the slope of the tangent line at x = 5. Since g(5) = 1:

    $\displaystyle g'(5) = \frac{1}{f'(1)}$

    Differentiate the given function:
    $\displaystyle f'(x) = 5x^4 + 9x^2 + 2$

    $\displaystyle f'(1) = 16$

    Thus:
    $\displaystyle g'(5) = \frac{1}{f'(1)} = \frac{1}{16} = 0.0625$
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