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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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You need to make $F$ continuous when $x = 1$. What is $F(1)$? What must $\lim_{x \rightarrow 1} F(x)$ equal in order for $F$ to be continuous at $x = 1$? To compute the limit, what are the one-sided limits, $\lim_{x \rightarrow 1^-} F(x)$ and $\lim_{x \rightarrow 1^+} F(x)$? Hope those hints help. – Michael Joyce Jan 17 '12 at 2:25
The only way to acknowledge the help of answerers in this forum is by accepting an answer. So, please consider doing so. – user21436 Jan 17 '12 at 2:37
Have you consider accept the answers? – leo Jan 17 '12 at 3:44
Please see this – leo Jan 17 '12 at 3:46

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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ok, so using what you told me, wouldn't a (in my problem) be 1? – Ronnie.j Jan 17 '12 at 2:48
@Ronnie.j: The typesetting is not in LaTeX, so I am not sure what the problem is. But if $f(x)=2x$ if $x\le 1$, and $f(x)=ax^2+1$ if $x>1$, then yes, $a=1$. What you wrote can also be interpreted as $f(x)=2x$ for $x\le 1$, $f(x)=2x+ax^2+1$ if $x>1$. If that's what you meant then $a=-1$. – André Nicolas Jan 17 '12 at 2:53

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