I was hoping someone could help me in finding a solution to a question regarding piecewise functions? I'm struggling because everything I have read / learned only has a single pronumeral and I can't find anything that contains multiple. The question is:
"Determine the values for the pronumerals that make the following piece-wise functions continuous."
$y = \begin{cases} 2x-1,\quad x\le4 \\ a+5x,\quad x>4 \end{cases}$
and
$y = \begin{cases} -x,\quad\quad x\le a \\ 2+x,\quad\ a<x\le b \\ 2x-5,\quad x>b \end{cases}$
Thanks in advance
Well it's nothing complicated. All you have to do is to remember that for a piecewise function to be continuous everywhere $$\lim_{x \to a^-}f(x) = \lim_{x \to a^+}f(x)$$ where $a$ is the point where the function changes/transforms into a different one.
Applying that logic in the first function, $$\lim_{x \to 4^-}f(x) = \lim_{x \to 4^+}f(x)$$ $$\implies \lim_{x \to 4^-}2x-1=\lim_{x \to 4^+} 5x+a$$ $$\implies 4 \times 2 - 1 = 5\times 4 +a$$ $$\implies a=-13$$
Hence for the first function $a=-13$.
Similarly for the second one try to find the points where the function changes into a different one and equate the limits to find the value of the pronumeral (Hint: The function changes twice in this case). Sometimes the piecewise function will also be told to be differentiable everywhere. In that case you may even need to find $f'(x)$ and equate it at the points where the derivative of the function changes.