1. Continuous Function

How to determine a and b so that function:

$
f(x) =
\begin{cases}
ax+b, & x\le -1, \\
4^x, & -1< x< 1, \\
ax+b, & 1\le x
\end{cases}
$

is continuous in all it's definition area?

2. hi

hi

$
f(x) =
\begin{cases}
ax+b, & x\le -1, \\
4^x, & -1< x< 1, \\
ax+b, & 1\le x
\end{cases}
$

Must have that $f(-1) = \frac{1}{4}$ , because
$4^{-1} = \frac{1}{4}$ .

And $f(1) = 4$

This gives $\begin{cases} -a +b = \frac{1}{4} \\ a + b = 4 \end{cases}$

$\Rightarrow \, a = \frac{15}{8} \mbox{ and } b = \frac{17}{8}$

3. Thank you for explanation.