Science:Math Exam Resources/Courses/MATH110/December 2015/Question 03 (a)

MATH110 December 2015

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Question 03 (a)
Consider the following function, $f(x)={\begin{cases}e^{x}&x\leq 0\\ax+b&0<x<1\\-x^{2}+x&x\geq 1.\end{cases}}$ where $a$ and $b$ are constants. (a) Find values of $a$ and $b$ for which $f$ is continuous everywhere.

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
A function $f$ is continuous at a point $a$ if if satisfies $f(a)=\lim _{x\to a+}f(x)=\lim _{x\to a-}f(x).$

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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. Recall that exponential functions and polynomials are continuous on the whole real line,. Therefore, it is enough to consider the continuity of function $f$ at $x=0$ and $x=1$. Using $\lim _{x\to 0-}f(x)=\lim _{x\to 0-}e^{x}=e^{0}=1=f(0)$ and $\lim _{x\to 0+}f(x)=\lim _{x\to 0+}ax+b=b$, to have the continuity at $x=0$, we need $b=1$. On the other hand, we have $\lim _{x\to 1-}f(x)=\lim _{x\to 1-}ax+b=a+b=a+1$ and $\lim _{x\to 1+}f(x)=\lim _{x\to 1+}-x^{2}+x=-(1^{2})+1=0=f(1)$. This implies that $f$ is continuous at $x=1$ when $a+1=0$. i.e., $a=-1$. To sum, the answers are $\color {blue}a=-1,b=1$.