Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (b)

MATH104 December 2012

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Question 01 (b)
For what value of the constant $a$ is the following function continuous at $x=1$? $f(x)={\begin{cases}{\frac {x\exp(x)-\exp(x)}{x^{2}-3x+2}},&x\neq 1\\a,&x=1\end{cases}}$

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Read the definition of continuity and check the conditions for a function to be continuous at a point. What value does $a$ need to be so that all the conditions for a function to be continuous at $x=1$ are satisfied? Also, the notation $\displaystyle \exp(x)$ is equivalent to $\displaystyle e^{x}$.

Hint 2
The conditions for a function $f(x)$ to be continuous at a point $x=c$ are: i. $\displaystyle f(c)\,\,{\mbox{exists}}$ ii. $\lim _{x\rightarrow c}f(x)\,\,{\mbox{exists}}$ iii. $\lim _{x\rightarrow c}f(x)=f(c)$.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. Following the second hint, we check all the conditions and see if there are any conditions where $a$ needs to take a particular value to satisfy the conditions: i. $\displaystyle f(1)=a.\quad$ So $f(1)$ exists. ii. To check that the limit at $x=1$ exists, we do not need to bother with left and right-hand limits as the function is defined in the same way on both sides of $x=1$. ${\begin{aligned}\lim _{x\rightarrow 1}{\frac {x\exp(x)-\exp(x)}{x^{2}-3x+2}}&=\lim _{x\rightarrow 1}{\frac {(x-1)\exp(x)}{(x-1)(x-2)}}\\&=\lim _{x\rightarrow 1}{\frac {\exp(x)}{(x-2)}}=-\exp(1)=-e\end{aligned}}$ So we have $\lim _{x\rightarrow 1}f(x)$ exists and is equal to $-e$. iii. The last condition to check is ${\begin{aligned}\lim _{x\rightarrow 1}f(x)&=f(1)\quad \rightarrow \quad -e=a\end{aligned}}$ So we can see that for the last condition of continuity to be satisfied we must have $a=-e$. Therefore, ${\color {blue}\displaystyle a=-e}$.

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