I am having a midterm review in school and there's one concept that I forgot how to solve, and that is solving for continuous functions?
More precisely, what does a variable have to be for the following to be continuous. For example, the problem I am dealt with solving is $$F(x)=\left\{\begin{array}{ll} 2x&\text{if }x\leq 1\\ ax^2+1&\text{if }x\gt 1\\ \end{array}\right.$$ and I have to solve for $a$. Normally, I would solve for $ax^2+1$, but I know that is wrong. Can someone tell me how to solve this, and perhaps by using a different problem so that I may be able to do the one I have on my own?
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$\begingroup$As per request, here is a problem that is like your problem.
Problem: Let $f(x)=5x$ when $x\le 2$, and let $f(x)=a^2x^2-7x$ when $x>2$. Find all values of $a$ such that $f$ is continuous everywhere.
Solution: Note that since $5x$ is continuous everywhere, $f$ is continuous at all $x<2$. Similarly, for any $a$, $a^2x^2-7x$ is continuous everywhere, so $f$ is continuous at all $x>2$. We now know that whatever choice we make for $a$, $f$ is continuous everywhere except possibly at $x=2$.
We want to find the values of $a$ such that $f$ is continuous at $x=2$.
As $x$ approaches $2$ from the left, $f(x)$ approaches $f(2)$. We want to make sure that as $x$ approaches $2$ from the right, $f(x)$ also approaches $f(2)$.
As $x$ approaches $2$ from the right, $f(x)$ approaches $a^2(2^2)-7(2)$. We want this "limit from the right" to be $f(2)$, that is, $10$. This will be the case precisely if $$4a^2 -14=10.$$ Solve for $a$. We get $a=\pm\sqrt{6}$.
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