Science:Math Exam Resources/Courses/MATH100/December 2013/Question 03 (c)

MATH100 December 2013

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Question 03 (c)
Short-Answer Question. Simplify your answer as much as possible and show your work. Let $\displaystyle f(x)={\begin{cases}cx^{2}+3&{\text{if }}x\geq 1\\2x^{3}-c&{\text{if }}x<1\end{cases}}$ where c is a constant. For what value of the constant c is the function ƒ(x) continuous on $\displaystyle (-\infty ,\infty )$?

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
What is the definition of continuity? At what $x$-value need we be concerned about the continuity of $f$?

Hint 2
A function is continuous at $\displaystyle a$ if $\lim _{x\to a}f(x)=f(a)$.

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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. We are concerned with the continuity of $f$ only at $x=1$, though we need not be concerned about whether $f(1)$ is defined or not, since its polynomial 'pieces' are defined on $(-\infty ,\infty )$. It remains to find a $c$ such that $\lim _{x\to 1}f(x)=\lim _{x\to 1^{-}}f(x)=\lim _{x\to 1^{+}}f(x)=f(1)$. By the definition of $\displaystyle f(x)$ we have $f(1)=\lim _{x\to 1^{+}}f(x)=\lim _{x\to 1^{+}}cx^{2}+3=c+3$. Furthermore, $\lim _{x\to 1^{-}}f(x)=\lim _{x\to 1^{-}}2x^{3}-c=2-c$. Hence we need to find a $\displaystyle c$ such that $\displaystyle 2-c=c+3$. It follows that the value $\displaystyle c$ that makes $\displaystyle f(x)$ continuous on $(-\infty ,\infty )$ is $c={\color {blue}{\frac {-1}{2}}}$.

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