## Calculus (3rd Edition)

Let $f(x)= x ^2.$ Note that $f$ is continuous on $\left[0 ,2\right]$ with $f(0)=0$ and $f(2)=4 .$ Therefore, by the IVT, there is a $c \in\left[0,2\right]$ such that $f(c)= c^2=2 .$ Thus the equation $x^2=2$ has a solution $c$ in $\left[0,2\right]$.