Science:Math Exam Resources/Courses/MATH100 C/December 2024/Question 01 (c)/Solution 1

Since $\underset{x\to c^{-}}{\lim }f(x)=c^2$ and $\underset{x\to c^{+}}{\lim }f(x)=6c$, we need to find values of $c$ for which $c^2=6c$. Therefore we obtain $c^2-6c=c(c-6)=0$, so $c=0$ or $c=6$.