guide to differentiability - algebraic way

• guide to limits
• guide to differentiation

This video explains an algebraic way to know if a function is differentiable at a point. here is the function described in the video:

Is $f(x)$ differentiable at $x=3$ for $f(x)=\{\begin{array}{ll}{x}^2& ,x<3\\ 6x-9& ,x\ge 3\end{array}$

We see that this is a piecewise function, and piecewise functions have a tendency to be, you know, having sharp turns.

We’ll think about this with algebra first, and then I’ll use a graphical interpretation to make things clear.

First and most importantly, continuity. Is it continuous? On one side we have the limit $(3{)}^2=9$ and on the other side we have $6(3)-9=9$. Looks like it’s continuous. But… is it a sharp turn? Is there an infinite slope?

The slope of $f(x)$ as $x\to 3^{-}$, is $\frac{d}{dx}[x^2]=2x=6$. What about as $x\to 3^{+}$? The slope is 6, the slopes are the same! The slope we approach from the left is the same as the slope we approach from the right; this means it’s differentiable.

Let’s do another example. We have this video now.

Is $f(x)$ differentiable at $x=1$ for $f(x)=\{\begin{array}{ll}x-1& ,x<1\\ (x-1{)}^2& ,x\ge 1\end{array}$

First, continuity. One one side we have $1-1=0$ and on the other side we have $(1-1{)}^2=0$. It’s continuous!
But what about sharp turns and infinite slope?
The slope of $f(x)$ as $x\to 1^{-}$, is $1$. What about as $x\to 1^{+}$? The slope is $\frac{d}{dx}[(x-1{)}^2]=(2)(x-1)(1)=2x-2$, the slopes are different! It is therefore a sharp turn, and the point is not differentiable.

Just one final example.
Is $f(x)$ differentiable at $x=4$ for $f(x)=\{\begin{array}{ll}2x+3& ,x<4\\ x^2-5& ,x\ge 4\end{array}$
First, continuity. Limit from the left is $2\cdot 4+3=11$ and right is $4^2-5=11$. It is continuous. What about slope? Slope from the left is $\frac{d}{dx}[2x+3]=2$, and slope from the right is $\frac{d}{dx}[x^2=5]=2x=8$. So not differentiable.

I’ll leave this up to you now… this exercise will help you a lot!

worked example: Is $f(x)$ differentiable at $x=2$ for $f(x)=\{\begin{array}{ll}{x}^2+1& ,x<2\\ x-x^2& ,x\ge 2\end{array}$
First continuity. One side has limit $(2{)}^2+1=5$ and the other has $2-(2{)}^2=-2$. It’s not continuous. And not differentiable.