Question

Determine the value of c so that f(x) is continuous on the entire real line when

Answer 1

\[\left\{ \left[\begin{matrix}x+3, & x \le-1 \\ 2x-c & x>-1\end{matrix}\right] \right\}\]

Answer 2

So we have a piece-wise function.
For differentiability, we need smoothness.
For continuity, we simply need connectedness.

We just need to find the correct constant value c, that will make the two pieces connect.

Answer 3

We will do this using limits.
Recall your limit stuff...
in order for a limit to exist, it's left and right sided limits must agree.

Answer 4

So we're going to look at the limit of our function from the left and right,
and force them to be equal to one another.

And we'll look for the c value that makes this happen.

Answer 5

\[\Large f(x)=\cases{x+3, &$x\le-1$\\ 2x-c, &$x\gt-1$}\]

We force the limits to agree, so that the limit will exist for -1,
\[\Large\rm \lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)\]

Answer 6

When we approach from the left side, we're using the top `piece`, yes?
See how that's only defined on the left side of -1?

Answer 7

yes

Answer 8

\[\Large\rm \lim_{x\to-1^-}x+3=\lim_{x\to-1^+}f(x)\]

Answer 9

And from the right side, when x is larger than -1, we use the other piece,\[\Large\rm \lim_{x\to-1^-}x+3=\lim_{x\to-1^+}2x-c\]

Answer 10

The process from here is actually really simple.
There is no discontinuity at -1 (at least there better not be),
so you just plug -1 directly in, then solve for c.

Answer 11

\[\Large\rm -1+3=2(-1)-c\]

Answer 12

2 = -2 - c
4 = - c
c = -4 ?

Answer 13

Mmm good good! :)

Here a graph of what that looks like:
https://www.desmos.com/calculator/rmoauzgoyu
Just in case you were curious.
The c=-4 gives us that connectedness that we needed.

Answer 14

Thanks a lot again.