Show that there is a number $c,$ with $0c1$ such that $f(c)=0.$

$f(x{)}^2=2x-60ix$

We have that $f(0)=1>0$ and $f(1)=2-cos1>0$ and that $\frac{?}{?}$ is continuous. Thus, by the inbermediate Value Theorem applied to $k=0.$ there is a number $c$ in $0,1$ such that $f(0)=a=0.$

$a=0,$ there in a number c in $||$ Q$.1||$ such that $f(c)-a=a.$

We have that $f(0)=-1<mn>0</mn>$ and $f(1)=2>0$ and that $f$ a continuous. Thas, by the intermediate Value Theorem applied to $k=$ Q$.$ there is a number $c$ in $0,1$ mach that $f()=k=0.$

We have that $f(0)=-1<mn>0</mn>$ and $f(1)=2-cos13-$ Oused sue $f$ inperiodic. Thas ty the leternediate Value Theorem applied to $k=0,$ there is a number $c$ in $0,1$ such that $f(s)=k=0.$

Theoren applied to $k=0,$ there is a namber $e$ in $0\mathrm{.}1$ such that $f(i)=a=0.$