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

$f(x)=2x-cosx$

We have that $f(0)=1>0$ and $f(1)=2-cos1>0$ and that $f$ is continuous. Thus, by the intermediate Vilue Theorem applied to $k=0,$ there is a number $c$ in $0,1$ such that $f(c)=k=0.$

We have that $f(0)=1>0$ and $f(1)=2>0$ and that $f$ is periodic. Thus, by the Intermediate Value Theorem applied to $k=0,$ there is a number $c$ in $0\mathrm{.}1$ such that $f(c)=k=0.$

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

We have that $f(0)=-1<mn>0</mn>$ and $f(1)=2-cos1>0$ and that $f$ is periodic. Thus, by the intermediate Value Theorem applied to $k=0,$ there is a number e in $0,1$ such that $f(c)=k=0.$

We have that $f(0)=-1<mn>0</mn>$ and $f(1)=2-cos1>0$ and that $f$ is continuous. Thus, by the Intermediate Value Theorem applied to $k=0,$ there h a number $c$ in $|0,1|$ such that $f(c)=k=0.$