# Question #f3eb0

Feb 8, 2018

$c = \frac{2}{3}$

#### Explanation:

For $f \left(x\right)$ to be continuous at $x = 2$, the following must be true:

• ${\lim}_{x \to 2} f \left(x\right)$ exists.
• $f \left(2\right)$ exists (this is not a problem here since $f \left(x\right)$ is clearly defined at $x = 2$

Let's investigate the first postulate. We know that for a limit to exist, the left hand and right hand limits must be equal. Mathematically:

${\lim}_{x \to {2}^{-}} f \left(x\right) = {\lim}_{x \to {2}^{+}} f \left(x\right)$

This also shows why we're only interested in $x = 2$: It's the only value of $x$ for which this function is defined as different things to the right and the left, which means that there is a chance the left & right hand limits may not be equal.

We'll be attempting to find values of 'c' for which these limits are equal.

Going back to the piecewise function, we see that to the left of $2$, $f \left(x\right) = c {x}^{2} + 2 x$. Alternatively, to the right of $x = 2$, we see that $f \left(x\right) = {x}^{3} - c x$

So:

${\lim}_{x \to 2} c {x}^{2} + 2 x = {\lim}_{x \to 2} {x}^{3} - c x$

Evaluating the limits:

${\left(2\right)}^{2} c + 2 \left(2\right) = {\left(2\right)}^{3} - \left(2\right) c$

$\implies 4 c + 4 = 8 - 2 c$

From here, it's just a matter of solving for $c$:

$6 c = 4$

$c = \frac{2}{3}$

What have we found? Well, we've figured out a value for $c$ that will make this function continuous everywhere. Any other value of $c$ and the right & left hand limits will not equal each other, and the function will not be continuous everywhere.

To get a visual idea of how this works, check out this interactive graph I made. Pick different values of $c$, and see how the function ceases to be continuous at $x = 2$!

Hope that helped :)