Consider the function defined by
\begin{align*} f(x) = \begin{cases} ax^{2} + x - b, & x < 2\\ ax+b, & 2\leq x \leq 5\\ 2ax - 7, & x > 5. \end{cases} \end{align*}
I was able to find a solution $(a,b)$ for the first two equations, but the solution would not work for the third. It would be amazing if someone could tell me how to find a solution for all three! Thank you!
$\endgroup$ 52 Answers
$\begingroup$First, remember the definition of continuity:
A function $f$ is continuous at $a$ ($a \in$ Dom$f$) if: $$\lim_{x \to a}f(x)=f(a)$$
Also, remember that a limit exists if and only if $$\lim_{x \to a^{-}}f(x) = \lim_{x \to a^{+}}f(x)$$
Now, you want $f$ to be continuous at $x=2$ and at $x=5$, using the preceding definitions at $x=2$:
$$\lim_{x \to 2^{-}}f(x)=\lim_{x \to 2^{+}}f(x)=\lim_{x \to 2}f(x)=f(2)$$
So, you have the following equality:
$$4a+2-b=2a+b \implies a=b-1$$
Doing the same for $x=5$:
$$\lim_{x \to 5^{-}}f(x)=\lim_{x \to 5^{+}}f(x)=\lim_{x \to 5}f(x)=f(5)$$
Then, evaluating the lateral limits, you have:
$$5a+b=10a-7 \implies 5a-7=b$$
Remember we obtained $a=b-1$, plugging in the last equation:
$$5a-7=b \implies 5(b-1)-7=b \implies b=3$$
Then $a=b-1 \implies a=3-1=2$.
Therefore, $a=2$ and $b=3$
$\endgroup$ $\begingroup$HINT\begin{align*} \begin{cases} \displaystyle f(2) = 2a + b = \lim_{x\rightarrow 2^{-}}f(x) = 4a + 2 -b\\\\ \displaystyle f(5) = 5a + b = \lim_{x\rightarrow 5^{+}}f(x) = 10a - 7 \end{cases} \end{align*}
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