Answer
Please see below.
Work Step by Step
(a)
Consider the function $g(x)= f_1(x)-f_2(x)$. Since $f_1(x)$ and $f_2(x)$ are continuous on the closed interval $[a,b]$, $g(x)$ is continuous on the closed interval $[a,b]$. Since $f_1(a) < f_2(a)$ and $f_1(b)> f_2(b)$, we have $g(a)=f_1(a)-f_2(a) < 0$ and $g(b)= f_1(b)-f_2(b)>0$. Thus, by Intermediate Value Theorem there must exist some real number $c$ between $a$ and $b$ such that $g(c)=f_1(c)-f_2(c) =0$; that is $f_1(c)=f_2(c)$.
(b)
Consider the functions $f_1(x)=x$ and $f_2(x)= \cos x$. These functions are continuous on the closed interval $[0, \frac {\pi }{2} ]$, and we have $f_1(0)=0 < f_2(0)=1$ and $f_1(\frac{\pi }{2})=\frac{\pi }{2} > f_2(\frac{\pi }{2})=0$. So, these functions satisfy the hypothesis of part (a). Thus, there exists some $c$ in $[0, \frac{\pi }{2} ]$ such that $\cos x =x$.
According to the graph, the intersection point of $x$ and $\cos x$ is approximately $c \approx 0.739$.
