Science:Math Exam Resources/Courses/MATH110/December 2013/Question 08/Solution 1

The condition for $f$ to be continuous is

${\begin{aligned}\lim _{x\rightarrow t^{-}}f(x)&=\lim _{x\rightarrow t^{+}}f(x)\\e^{-t}&=2(t+1)\end{aligned}}$

Since both piece of the function are continuous, we can use direct substitution and find $t$. We must show there exists a solution to the above equation. Equivalently we want to show that the following function $g(t)$ has a zero

${\begin{aligned}g(t)=e^{-t}-2(t+1)\end{aligned}}$

Note that $g$ is continuous since it is constructed from continuous functions. We aim to apply IVT. Observe

${\begin{aligned}g(0)&=e^{0}-2<0\\g(-1)&=e^{1}-2(-1+1)=e>0\end{aligned}}$

Hence by the IVT there exists $c\in (-1,0)$ such that $g(c)=0$. At $t=c$, we have $e^{-c}=2(c+1)$. In turn, this means we have the two one sided limits equalling each other and hence $f(x)$ will be continuous by choosing $x=c$.