Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (d)/Solution 1

Our goal is to find f(3). Since f(x) is continuous we know that $f(3)=\lim _{x\to 3}f(x)$ and thus finding f(3) amounts to finding the limit.

Using the limit laws we have

$1=\lim _{x\to 3}(xf(x)+g(x))=\lim _{x\to 3}x\lim _{x\to 3}f(x)+\lim _{x\to 3}g(x).$

Using that both functions x and g(x) are continuous, we have that

$\lim _{x\to 3}x=3$

and

$\lim _{x\to 3}g(x)=g(3)=2.$

Plugin this into the above equality, we find

$1=3\lim _{x\to 3}f(x)+2$

and hence

$\lim _{x\to 3}f(x)=-{\frac {1}{3}}.$

By the first remark above we have

$\displaystyle f(3)=-{\frac {1}{3}}.$