# Evaluating differentiability, and continuity of a piecewise defined function

$f(x) = \begin{cases} x^2-8, & x\lt 3 \\ 1+5\ln(x-2), & x \ge 3 \end{cases}$

First I needed to determine if $f$ is continuous at 3. So I plugged in $3$ to both of pieces of $f$ and both of them are $1$, which means it is continuous at 3.

Next I needed to determine the limit of:

$\lim \limits_{x \to 3^-}\frac{f(x)-f(3)}{x-3}=6$

And:

$\lim \limits_{x \to 3^+}\frac{f(x)-f(3)}{x-3}=5$

So the next question is is it differentiable at $3$. My answer is yes because I can clearly take the derivative. So while all of the other things I mentioned could be incorrect, the follow up question has me very confused, it asks if $f$ is differentiable at $3$ then what is $f'(3)$?

This all caused me to go and re-read the definition for a continuous function and a differentiable function and wiki says the following:

If $f$ is differentiable at a point $x_0$, then f must also be continuous at $x_0$.

It seems like I meet those requirements.

• You need to use the definition of continuity and differentiability, which involve the existence of certain limits. – mathematician Jul 23 '16 at 1:27
• @mathematician right... I'm trying to apply them but I'm confused by their application to this problem. For example, if I graph these two functions they have an intersection at $y=1$, there aren't any holes or jumps but they split from one another. So since there are no holes or jumps, it would be continuous at this point it seems. – hax0r_n_code Jul 23 '16 at 1:29
• If A implies B, then B does not have to imply A – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Jul 23 '16 at 1:37

You have actually proven that $$f'(3) = \lim_{x \to 3} \frac{f(x)-f(3)}{x-3}$$ doesn't exist, as it takes two distinct values if you approach from the left and from the right. So this is an example of a function which is continuous at a point but not differentiable there. (As another example, consider $f(x) = \lvert x \rvert$ at $x=0$.)
As you can see from this zoomed-in plot of the function, the "lines" meeting at $x = 3$ don't have the same slope! As you computed, the one on the left has a slope of $6$, but the one on the right has a slope of $5$.
• How is it that taking the limit for each part of the piecewise function is equal to $1$? What does this tell me? Sorry I'm slightly confused still – hax0r_n_code Jul 23 '16 at 1:37