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Mathematics LibreTexts

8.5: The Exponential Function

  • Page ID
    22689
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    Definition

    We call the inverse of the logarithm function the exponential function. We denote the value of the exponential function at a real number \(x\) by \(\exp (x)\).

    Proposition \(\PageIndex{1}\)

    The exponential function has domain \(\mathrm{R}\) and range \((0,+\infty)\). Moreover, the exponential function is increasing and differentiable on \(\mathbb{R}\). If \(f(x)=\exp (x),\) then \(f^{\prime}(x)=\exp (x)\).

    Proof

    Only the final statement of the proposition requires proof. If we let \(g(x)=\log (x),\) then

    \[f^{\prime}(x)=\frac{1}{g^{\prime}(\exp (x))}=\exp (x).\]

    Q.E.D.

    Proposition \(\PageIndex{2}\)

    For any real numbers \(x\) and \(y\),

    \[\exp (x+y)=\exp (x) \exp (y).\]

    Proof

    The result follows from

    \[\log (\exp (x) \exp (y))=\log (\exp (x))+\log (\exp (y))=x+y.\]

    Q.E.D.

    Proposition \(\PageIndex{3}\)

    For any real number \(x\),

    \[\exp (-x)=\frac{1}{\exp (x)}.\]

    Proof

    The result follows from

    \[\log \left(\frac{1}{\exp (x)}\right)=-\log (\exp (x))=-x.\]

    Q.E.D.

    Exercise \(\PageIndex{1}\)

    Use Thylor's theorem to show that

    \[\exp (1)=e=\sum_{n=0}^{\infty} \frac{1}{n !}.\]

    Proposition \(\PageIndex{4}\)

    For any rational number \(\alpha\),

    \[\exp (\alpha)=e^{\alpha}.\]

    Proof

    Since \(\log (e)=1,\) we have

    \[\log \left(e^{\alpha}\right)=\alpha \log (e)=\alpha.\]

    Q.E.D.

    Definition

    If \(\alpha\) is an irrational number, we define

    \[e^{\alpha}=\exp (\alpha).\]

    Note that for any real numbers \(x\) and \(y\),

    \[e^{x+y}=e^{x} e^{y}\]

    and

    \[e^{-x}=\frac{1}{e^{x}}.\]

    Moreover, \(\log \left(e^{x}\right)=x\) and, if \(x>0, e^{\log (x)}=x\).

    Definition

    If \(x\) and \(a\) are real numbers with \(a>0,\) we define

    \[a^{x}=e^{x \log (a)}.\]

    Exercise \(\PageIndex{2}\)

    Define \(f:(0,+\infty) \rightarrow \mathbb{R}\) by \(f(x)=x^{a},\) where \(a \in \mathbb{R}, a \neq 0\). Show that \(f^{\prime}(x)=a x^{a-1}\).

    Exercise \(\PageIndex{3}\)

    Suppose \(a\) is a positive real number and \(f: \mathbb{R} \rightarrow \mathbb{R}\) is defined by \(f(x)=a^{x} .\) Show that \(f^{\prime}(x)=a^{x} \log (a)\).

    Proposition \(\PageIndex{5}\)

    For any real number \(\alpha>0\),

    \[\lim _{x \rightarrow+\infty} x^{\alpha} e^{-x}=0.\]

    Proof

    We know that

    \[\lim _{y \rightarrow+\infty} \frac{\log (y)}{y^{\frac{1}{2}}}=0.\]

    Hence

    \[\lim _{y \rightarrow+\infty} \frac{(\log (y))^{\alpha}}{y}=0.\]

    Letting \(y=e^{x},\) we have

    \[\lim _{x \rightarrow+\infty} \frac{x^{\alpha}}{e^{x}}=0.\]

    Q.E.D.

    Proposition \(\PageIndex{6}\)

    For any real number \(\alpha\),

    \[\lim _{x \rightarrow+\infty}\left(1+\frac{\alpha}{x}\right)^{x}=e^{\alpha}.\]

    Proof

    First note that, letting \(x=\frac{1}{h}\),

    \[\lim _{x \rightarrow+\infty}\left(1+\frac{\alpha}{x}\right)^{x}=\lim _{h \rightarrow 0+}(1+\alpha h)^{\frac{1}{k}}=\lim _{h \rightarrow 0^{+}} e^{\frac{1}{h} \log (1+\alpha h)}.\]

    Using I'Hópital's rule, we have

    \[\lim _{h \rightarrow 0^{+}} \frac{\log (1+\alpha h)}{h}=\lim _{h \rightarrow 0^{+}} \frac{\alpha}{1+\alpha h}=\alpha,\]

    and the result follows from the continuity of the exponential function. \(\quad\) Q.E.D.

    Definition

    We define the hyperbolic sine and hyperbolic cosine functions by

    \[\sinh (x)=\frac{e^{x}-e^{-x}}{2}\]

    and

    \[\cosh (x)=\frac{e^{x}+e^{-x}}{2},\]

    respectively.

    Exercise \(\PageIndex{4}\)

    Show that for any real numbers \(x\) and \(y\),

    \[\sinh (x+y)=\sinh (x) \cosh (y)+\sinh (y) \cosh (x)\]

    and

    \[\cosh (x+y)=\cosh (x) \cosh (y)+\sinh (x) \sinh (y).\]

    Exercise \(\PageIndex{5}\)

    Show that for any real number \(x\),

    \[\cosh ^{2}(x)-\sinh ^{2}(x)=1.\]

    Exercise \(\PageIndex{6}\)

    If \(f(x)=\sinh (x)\) and \(g(x)=\cosh (x),\) show that

    \[f^{\prime}(x)=\cosh (x)\]

    and

    \[g^{\prime}(x)=\sinh (x).\]

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