Show that the equation $x^3-14xc=0$ has at most one solution in the interval $-2,2.$

Let $f(x)=x^3-14xc$ for $x$ in $-2,2,$ If $f$ has two real solutions a and $b$ in $-2,2,$ with then $--$latesSince the polynomiol $fis$ continuous $ona,b$ and offerentable $on(a,b).--$Select$-mq,Q$ implies that there $is$ a number $rin(a,b)$ such that $r^{\text{'}}(r)=0.$

Now $f^{\text{'}}(r)=-$ Since $risin(a,b),$ which $is$ contained $in-2,2,we$ have follows that $3r^2-14=83-4-14=-200.$ This contraticts $r||$