Question

Determine the values of a and b for which the following function is continuous at every real number x

f(x) { ax+b if x ≤ 1
-2bx^2 - 8a if 1 < x < 2
4b^2/a -5x if x ≥ 2 }

Answer 1

continuous implies \[\lim_{x \rightarrow c}f(x)=f(c)\]

for the first one, \[\lim_{x \rightarrow 1}f(x)=f(1)\]

\[\lim_{x \rightarrow 1}(ax+b)=a+b\] and so on...

Answer 2

\[f(x)=\begin{cases}ax+b&\text{if }x\le1\\
-2bx^2-8a&\text{if }1<x<2\\
\frac{4b^2}{a-5x}&\text{if }x\ge2\end{cases}\]
For f(x) to be continuous at x = 1 and x = 2, the following must be satisfied:
\[\color{red}{\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)}\\
\color{blue}{\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)}\]
\[RED:\lim_{x\to1^-}(ax+b)=\lim_{x\to1^+}(-2bx^2-8a)\\
BLUE: \lim_{x\to2^-}(-2bx^2-8a)=\lim_{x\to2^+}\frac{4b^2}{a-5x}\]
Determining the limits should give you a system of equations (assuming I rewrote the function you provided).