3 - Ximera

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We see that if a function is differentiable at a point, then it must be continuous at that point.

There are connections between continuity and differentiability.

Differentiability Implies Continuity If $f$ is a differentiable function at $x = a$ , then $f$ is continuous at $x=a$ .

To explain why this is true, we are going to use the following definition of the derivative $f'(a) = \lim _{x\to a} \frac {f(x)-f(a)}{x-a}.$

Assuming that $f'(a)$ exists, we want to show that $f(x)$ is continuous at $x=a$ , hence we must show that $\lim _{x\to a} f(x) = f(a).$ Starting with $\lim _{x\to a} \left (f(x) - f(a)\right )$ we multiply and divide by $(x-a)$ to get

$\begin{align*} &= \lim_{x\to a} \left((x-a)\frac{f(x) - f(a)}{x-a}\right) \\ &= \left(\lim_{x\to a} (x-a) \right) \left(\lim_{x\to a}\frac{f(x) - f(a)}{x-a}\right) &\text{Limit Law.} \\ &= \answer[given]{0}\cdot f'(a) = \answer[given]{0}. \end{align*}$

Since $\lim _{x\to a}\left (f(x) - f(a)\right ) = 0$ we see that $\lim _{x\to a} f(x) = f(a)$ , and so $f$ is continuous at $x=a$ .

This theorem is often written as its contrapositive:

If $f(x)$ is not continuous at $x=a$ , then $f(x)$ is not differentiable at $x=a$ .

Thus from the theorem above, we see that all differentiable functions on $\RR$ are continuous on $\RR$ . Nevertheless there are continuous functions on $\RR$ that are not differentiable on $\RR$ .

Which of the following functions are continuous but not differentiable on $\RR$ ?

$x^2$ $\lfloor x \rfloor$ $|x|$ $\frac {e^{x}}{x}$

Consider $f(x) = \begin {cases} x^2 &\text {if $x<3$,}\\ mx+b &\text {if $x\ge 3$.} \end {cases}$ What values of $m$ and $b$ make $f$ differentiable at $x=3$ ?

To start, we know that we must make $f$ both continuous and differentiable. We will start by showing $f$ is continuous at $x=3$ . Write with me: $\begin{align*} \lim_{x\to 3^-} f(x) &= \answer[given]{9}\\ \lim_{x\to 3^+} f(x) &= \answer[given]{m 3 + b}\\ f(3) &= \answer[given]{m 3 + b} \end{align*}$

So for the function to be continuous, we must have $m\cdot 3 + b =9.$ We also must ensure that the value of the derivatives of both pieces of $f$ agree at $x=3$ . Write with me

$\begin{align*} \ddx x^2 &= \lim_{h\to 0}\frac{(x+h)^2-x^2}{h}\\ &= \lim_{h\to 0}\frac{x^2 + 2xh + h^2 - x^2}{h}\\ &= \lim_{h\to 0}\frac{2xh + h^2}{h}\\ &= \lim_{h\to 0}\left(2x + h\right)\\ &=2x. \end{align*}$

Moreover, by the definition of a tangent line $\ddx (mx+ b) = \answer [given]{m}$ Hence we must have

$\begin{align*} \eval{\ddx x^2}_{x=3} &= \eval{\ddx(mx+b)}_{x=3}\\ \eval{2x}_{x=3} &= \eval{m}_{x=3}\\ 6 &= m. \end{align*}$

Ah! So now

$\begin{align*} 9 &= m\cdot 3 + b\\ 9 &= 6\cdot 3 + b\\ 9 &= 18 + b, \end{align*}$

so $b=\answer [given]{-9}$ . Thus setting $m=\answer [given]{6}$ and $b=\answer [given]{-9}$ will give us a differentiable (and hence continuous) piecewise function. We can confirm our results by looking at the graph of $y=f(x)$ : $\graph [xmin=0,xmax=6,ymin=-1,ymax=15]{y=x^2\left \{x<3\right \},y=mx+b\left \{3 \leq x\right \},m=6,b=-9}$