If I have a piecewise function, must I prove it is continous to show it is differentiable at a point? Or is it if I am able to apply the derivative rules to the function, it must be continous and therefore differentiable?
For Example:Is $f\left(x\right)$ is differentiable at 2? $$ f(x) = \left\{ \begin{array}{ll} 4x^2+1 & \quad x > 2 \\ 17 & \quad x = 2 \\ 16x-15 & \quad x < 2 \end{array} \right. $$ Do I have to show $\lim \:_{x\to \:\:2}\:f\left(x\right)=f\left(2\right)$? Or can I assume that is true if $\frac{d}{dx+2}\left(f\left(x\right)\right)=\frac{d}{dx-2}\left(f\left(x\right)\right)$? I guess what I am really asking what is the criteria for a piece wise function to be differentiable at a point?
$$ f'(x) = \left\{ \begin{array}{ll} 8x & \quad x > 2 \\ 0 & \quad x = 2 \\ 16 & \quad x < 2 \end{array} \right. $$
Is showing $f'_+\left(2\right)=f'_-\left(2\right)=16$ enough to say this function is differentiable? Also does it matter that $f'\left(2\right)=0$ and not $16$.
2 Answers
To prove that a function is differentiable at a point $x \in \mathbb{R}$ we must prove that the limit $$ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ exists. As an example let us study the differentiability of your function at $x = 2$ we have $$ \frac{f(2 + h) - f(2)}{2} = \frac{f(2 + h) - 17}{h}$$ Now if $h > 0$ we have the right-side limit $$ \lim_{h \to 0^+} \frac{4(2 + h)^2 + 1 - 17}{h} = \lim_{h \to 0^+} \frac{16 + 16h + 4h^2 + 1 - 17}{h} = \lim_{h \to 0^+} \frac{16h + 4h^2}{h} = 16$$ if $h < 0$ we have the left-side limit $$ \lim_{h \to 0^-} \frac{16(2 + h) - 15 - 17}{h} = \lim_{h \to 0^-} \frac{32 + 16h - 15 - 17}{h} = 16$$ So the left and right sided limits agree, and the limit exists. We may conclude that $f$ is differentiable at $x = 2$. Can you handle the other point?
An additional note on your answer Consider carefully whether your derivative "function" makes sense. You have implicitly assumed that $f'(2) = 0$ without proof! Note that in your version of the derivative function if $f(x) = a \in \mathbb{R}$ for some $x \in \mathbb{R}$ then $f'(x) = 0$. Consider the function defined by $$ f(x) = \begin{cases}2x, \ x > \frac{1}{2} \\ 1, \ x = \frac{1}{2} \\ 2x, \ x < \frac{1}{2} \end{cases}$$ This is a roundabout way to define $f(x) = 2x$. Using your formulation we would have $$ f'(x) = \begin{cases}2, \ x > \frac{1}{2} \\ 0, \ x = \frac{1}{2} \\ 2, \ x < \frac{1}{2} \end{cases}$$ which is clearly absurd!
A function $f$ is said to be continuous at a point $a$ if \begin{equation*} \lim_{x\rightarrow a} f(x) = f(a) \end{equation*}
A function $f$ is said to be differentiable at a point $a$ if \begin{equation*} \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a} \qquad \text{exists} \end{equation*}
So differentiable implies continuous (prove it as an exercise if you want; so you need to verify that
\begin{equation*} \lim_{x \rightarrow 2} \frac{f(x) f(2)}{x-2} \end{equation*} exists, and you need to analyze the two possible ways that $x$ can tend to $2$ (left or right).
