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QUESTION

\begin{description}

\item[(i)]
On a single set of axis sketch the graphs of $y=\cosh x$ and
$y=\cosh^{-1}x$. State the domains and ranges of these functions.

\item[(ii)]
If $y=\cosh^{-1}x$ use the exponential definition of $\cosh y$ to
show that y satisfies the equation

$$e^{2y}-2xe^y+1=0.$$

\item[(iii)]
Rewrite the above equation as a quadratic and hence deduce that

$$\cosh^{-1}x=\ln\{x+\sqrt{x^2-1}\}.$$

\item[(iv)]
By differentiating the result in (iii) verify that

$$\frac{d}{dx}(\cosh^{-1}x)=\frac{1}{\sqrt{x^2-1}}.$$

\end{description}


ANSWER

\begin{description}

\item[(i)]

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\epsfig{file=cosh01.eps, width=70mm}
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For $\cosh:$ the domain is $-\infty<x<\infty$\hspace{1cm}The range
is $y\geq1.$

For $\cosh^{-1}:$ the domain is $x\ge1$\hspace{1cm} The range is
 $y\geq0.$

\item[(ii)]
$y=\cosh^{-1}x$ therefore $x=\cosh y$\\ Thus $\ds
x=\frac{e^y+e^{-y}}{2}$ i.e. $2xe^y=e^{2y}+1$ or
$e^{2y}-2xe^y+1=0$

\item[(iii)]
Hence $(e^y)^2-2xe^y+1=0$ which is a quadratic in $e^y$ so

\begin{eqnarray*}
e^y&=&\frac{2x\pm\sqrt{((-2x)^2-4(1)(1))}}{2}\\
&=&\frac{2x\pm\sqrt{4x^2-4}}{2}\\ e^y&=&x\pm\sqrt{x^2-1}\\
\Rightarrow y&=&\ln\{x\pm\sqrt{x^2-1}\}
\end{eqnarray*}

Now
\begin{eqnarray*}
&&\ln\{x+\sqrt{x^2-1}\}+\ln\{x-\sqrt{x^2-1}\}\\
&&=\ln\{(x+\sqrt{x^2-1})(x-\sqrt{x^2-1})\}\\
&&=\ln\{x^2-(x^2-1)\}\\ &&=\ln1=0\\\textrm{Therefore}\\
\ln\{x-\sqrt{x^2-1}\}&&=-\ln\{x+\sqrt{x^2-1}\}\\\textrm{ so }
y&&=\pm\ln\{x+\sqrt{x^2-1}\}
\end{eqnarray*}

When $x\geq1,\
x+\sqrt{x^2+1}\geq1\Rightarrow\ln\{x+\sqrt{x^2-1}\}\geq0$\\
Therefore for the inverse, $\cosh^{-1}x=\ln\{x+\sqrt{x^2-1}\}$

\item[(iv)]

\begin{eqnarray*}
\frac{d}{dx}\left\{\cosh^{-1}x\right\}&=&
\frac{1}{x+\sqrt{x^2-1}}\times\left\{1+\frac{1}{2}(x^2-1)^{-\frac{1}{2}}(2x)\right\}\\
&=&\frac{1}{x+\sqrt{x^2-1}}\left\{1+\frac{x}{\sqrt{x^2-1}}\right\}\\
&=&\frac{\sqrt{x^2-1}+x}{\left\{x+\sqrt{x^2-1}\right\}\sqrt{x^2-1}}\\
&=&\frac{1}{\sqrt{x^2-1}}
\end{eqnarray*}

\end{description}



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