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

MATH110 December 2013

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Question 08
Let $f(x)=\left\{{\begin{array}{lr}e^{-x}&{\text{if}}\quad x<t\\2(x+1)&{\text{if}}\quad x\geq t\end{array}}\right.$ Explain why there exists a number $t$ such that $f$ is continuous. (Hint: apply the Intermediate Value Theorem to the function $e^{-t}-2(t+1)$.)

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
We write down what it means for $f$ to be continuous. This is given by ${\begin{aligned}\lim _{x\rightarrow t^{-}}f(x)&=\lim _{x\rightarrow t^{+}}f(x)\\e^{-t}&=2(t+1)\end{aligned}}$ In order for our function to be continuous, we need a solution to the above equation. Now we can apply the hint in the problem to use the intermediate value theorem.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $x=c$.

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