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

The condition for $f$ to be continuous is

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

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 $x = c$ .