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Mathematics LibreTexts

4.4: Inverse Functions

  • Page ID
    528
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    An inverse function is a function that undoes another function: If an input \(x\) into the function \(f\) produces an output \(y\), then putting \(y\) into the inverse function \(g\) produces the output \(x\), and vice versa.

    Definition: Inverse Functions

    Let \(f(x)\) be a 1-1 function then \(g(x)\) is an inverse function of \(f(x)\) if

    \[ f(g(x)) = g(f(x)) = x. \]

    Example \(\PageIndex{1}\)

    For

    \[ f(x) = 2x - 1 \nonumber \]

    \[ f^{ -1}(x) = \dfrac{1}{2} x +\dfrac{1}{2} \nonumber \]

    since

    \[ f(f^{ -1}(x) ) = 2[\dfrac{1}{2} x +\dfrac{1}{2}] - 1 = x \nonumber \]

    and

    \[ f ^{-1}(f(x)) = \dfrac{1}{2} [2x - 1] + \dfrac{1}{2} = x. \nonumber \]

    The Horizontal Line Test and Roll's Theorem

    Note that if \(f(x)\) is differentiable and the horizontal line test fails then

    \[ f(a) = f(b) \]

    and Rolls theorem implies that there is a \(c\) such that

    \[ f '(c) = 0.\]

    A partial converse is also true:

    Theorem (Roll's Theorem)

    If \(f\) is differentiable and \(f '(x)\) is always non-negative (or always non-positive) then \(f(x)\) has an inverse.

    Example \(\PageIndex{2}\)

    \[ f(x) = x^3 + x - 4 \nonumber \]

    has an inverse since

    \[f'(x) = 3x^2 + 1 \nonumber \]

    which is always positive.

    Continuity and Differentiability of the Inverse Function

    Theorem (Continuity and Differentiability)

    1. \(f\) continuous implies that \(f^{ -1}\) is continuous.
    2. \(f\) increasing implies that \(f^{ -1}\) is increasing.
    3. \(f\) decreasing implies that \(f^{ -1}\) is decreasing.
    4. \(f\) differentiable at \(c\) and \(f '(c) \neq 0\) implies that \(f^{ -1}\) is differentiable at \(f (c)\).
    5. If \(g(x)\) is the inverse of the differentiable \(f(x)\) then 

    \[ g'(x) = \dfrac{1}{f '(g(x))}.\]

    if \(f '(g(x)) \neq 0\).

    Proof

    Proof of (5)

    Since

    \[ f (g(x)) = x \nonumber \]

    we differentiate implicitly:

    \[\dfrac{d}{dx} f(g(x)) = \dfrac{d}{dx} x.\nonumber \]

    Using the chain rule

    \[ y =f(u), u = g(x)\nonumber \]

    \[ \begin{align*} \dfrac{dy}{x} &= \dfrac{dy}{dy} \dfrac{du}{dx} \nonumber \\[5pt] &= f '(u) g'(x) = f '(g(x)) g'(x). \end{align*}\]

    So that

    \[ f '(g(x)) g'(x) = 1. \nonumber \]

    Dividing, we get:

    \[g'(x) = \dfrac{1}{f'(g(x))}.\nonumber \]

    \(\square\)

    Example \(\PageIndex{3}\)

    For \(x > 0\), let

    \[ f(x) = x^2\nonumber \]

    and

    \[ g(x) = \sqrt{x}\nonumber \]

    be its inverse, then

    \[ g'(x) = \dfrac{1}{2\sqrt{x}}.\nonumber \]

    Note that 

    \[ \dfrac{d}{dx} \sqrt{x} = \dfrac{d}{dx} x^{\frac{1}{2}} = \dfrac{1}{2} x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}.\nonumber \]

    Exercises

    1. Let

    \[ f(x) = x^3 + x - 4.\nonumber \]

    Find

    \[ \dfrac{d}{dx} f^{-1}(-4). \nonumber \]

    2. Let 

    \[f(x) =\int_2^x \dfrac{1}{1+x^3} dx\nonumber \]

    Find 

    \[ \dfrac{d}{dx} f^{-1}(0). \nonumber \]

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