Let \(f(x) = \begin{cases} 3x^2 + 2&\text{if } x\le 3, \\ ax - 1 &\text{if } x>3. \end{cases} \) find \(a\) if the graph of \(y=f(x)\) is continuous (which means the graph can be drawn without lifting your pencil from the paper).
+2446 I am not sure how Guest arrived at a=5, but that answer is incorrect.
In order for a function to be continuous at x=c, the function must fulfill three criteria. They are the following:
1) \(f(c)\) is defined.
2) \(\lim_{x\rightarrow c}f(x)\) is defined.
3) \(\lim_{x\rightarrow c}f(x)=f(c)\)
Right now, only one criterion is met currently. \(f(3)\) is defined in this piecewise function because \(x=3\) is included in the domain. Let's move on to the next condition.
In order for \(\lim_{x\rightarrow c}f(x)\) to be defined, we have to ensure that left-side and right-side limit approach the same value. We set both limits equal to each other and find possible values for a, if one exists.
\(\lim_{x\rightarrow 3^-}f(x)=\lim_{x\rightarrow 3^+}f(x)\\ 3*3^2+2=3a-1\\ 29=3a-1\\ 3a=30\\ a=10 \)
At \(a=10\), both limits approach the same value. Therefore, we have met the second condition.
Finally, we have to make sure that the limit and the true value are not incongruous.
\(\lim_{x\rightarrow 3}f(x)\stackrel{?}{=}f(3)\\ 29=29\)
Yes, that means that a=10 makes the function continuous.
