49/65

\begin{frame} \frametitle{1st Midterm Exam - Review} \begin{exampleblock}{} Prove that the equation \begin{talign} 2\cos(\frac{x\pi}{2}) = 2^x \end{talign} has a solution for $x$ on the interval $[0,1]$. \pause\medskip Define $f(x) = 2\cos(\frac{x\pi}{2}) - 2^x$: \begin{talign} 2\cos(\frac{x\pi}{2}) = 2^x &&\iff&& f(x) = 0 \end{talign} \pause We have: \begin{itemize} \pause \item $f(x)$ is defined on $[0,1]$ \pause \item $f(x)$ is continuous on its domain since it is a composition, product, division and multiplication of continuous functions \pause \item $f(0) = \pause 1$ \pause and $f(1) = \pause -2$ \end{itemize} \pause Since $0$ is between $-2$ and $1$, by the Intermediate Value Theorem, \pause there exists $c$ in $(0,1)$ such that $f(c) = 0$. \pause This $c$ is a solution of the equation. \end{exampleblock} \end{frame}