Differentiability Implies Continuity

An important theorem concerning derivatives is this:

If a function f is differentiable at x = a, then f is continuous at x = a.

The proof begins with the identity that for all $x\ne a$

$\displaystyle f\left( x \right)-f\left( a \right)=\left( {x-a} \right)\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}$

$\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left( {f\left( x \right)-f\left( a \right)} \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {\left( {x-a} \right)\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}} \right)=\underset{{x\to a}}{\mathop{{\lim }}}\,\left( {x-a} \right)\cdot \underset{{x\to a}}{\mathop{{\lim }}}\,\frac{{f\left( x \right)-f\left( a \right)}}{{x-a}}$

$\displaystyle \underset{{x\to a}}{\mathop{{\lim }}}\,\left( {f\left( x \right)-f\left( a \right)} \right)=0\cdot {f}'\left( a \right)=0$

And therefore, $\underset{{x\to a}}{\mathop{{\lim }}}\,f\left( x \right)=f\left( a \right)$

Since both sides are finite, the function is continuous at x = a.

The converse of this theorem is false: A continuous function is not necessarily differentiable. A counterexample is the absolute value function which is continuous at the origin but not differentiable there. (The slope approaching from the left is not equal to the slope from the right.)

This is a theorem whose contrapositive is used as much as the theorem itself. The contrapositive is,

If a function is not continuous at a point, then it is not differentiable there.

Example 1: A function such as $\displaystyle g\left( x \right)=\frac{{{{x}^{2}}-9}}{{x-3}}$ has a (removable) discontinuity at x = 3, but no value there.

So, in the limit definition of the derivative, $\displaystyle \text{ }\!\!~\!\!\text{ }\underset{{h\to 0}}{\mathop{{\lim }}}\,\frac{{g\left( {3+h} \right)-g\left( 3 \right)}}{h}$ there is no value of g(3) to use, and the derivative does not exist.

Example 2: $\displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}} & {x\le 1} \\ {{{x}^{2}}+3} & {x>1} \end{array}} \right.$ . This function has a jump discontinuity at x = 1.

Since the point (1, 1) is on the left part of the graph, if h > 0, $f\left( {1+h} \right)-f\left( 1 \right)>3$ and the limit will always be a number greater than 3 divided by zero and will not exist. Therefore, even though the slopes from both side of x =1 approach the same value, namely 2, the derivative does not exist at x = 1.

This also applies to a situation like example 1 if f(3) were some value that did not fill in the hole in the graph.

On the AP Calculus exams students are often asked about the derivative of a function like those in the examples, and the lack of continuity should be an immediate clue that the derivative does not exist. See 2008 AB 6 (multiple-choice).

Just as important are questions in which the function is given as differentiable, but the student needs to know about continuity. Just remember: differentiability implies continuity. See 2013 AB 14 in which you must realize the since the function is given as differentiable at x = 1, it must be continuous there to solve the problem.

Continuity of the Derivative

A question that comes up is, if a function is differentiable is its derivative differentiable? The answer is no. While almost always the derivative is also differentiable, there is this counterexample:

$\displaystyle f\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {{{x}^{2}}\sin \left( {\frac{1}{x}} \right)} & {x\ne 0} \\ 0 & {x=0} \end{array}} \right.$

The first line of the function has a removable oscillating discontinuity at x = 0, but since the $\displaystyle {{x}^{2}}$ factor squeezes the function to the origin; the added condition that $\displaystyle f\left( 0 \right)=0$ makes the function continuous. Differentiating gives

$\displaystyle {{f}^{'}}\left( x \right)={{x}^{2}}\cos \left( {\frac{1}{x}} \right)\left( {\frac{{-1}}{{{{x}^{2}}}}} \right)+2x\sin \left( {\frac{1}{x}} \right)=-\cos \left( {\frac{1}{x}} \right)+2x\sin \left( {\frac{1}{x}} \right)$

And now there is no way to get around the oscillating discontinuity at x = 0.

Teaching Calculus

The pleasure lies not in discovering the truth, but in searching for it.

Differentiability Implies Continuity

Leave a Reply Cancel reply

Share this:

Like this:

Related

Leave a Reply Cancel reply