Associated Topics || Dr. Math Home || Search Dr. Math

### Continuity and Differentiability of Piecewise Function

```
Date: 05/04/98 at 11:25:23
From: Heather Hopson
Subject: differentiable and continous functions

A function is defined by the following formula:

f(x) = x^2 + 2, for x less than or equal to 1
= a(x - (1/x)) + b, for x greater than 1

Find a and b such that f is continuous and differentiable. Plot the
function, if possible.

How do I find a and b? If it is continuous, isn't it differentiable?
```

```
Date: 05/04/98 at 15:53:39
From: Doctor Sam
Subject: Re: differentiable and continous functions

Heather,

function can be continuous without having a derivative. The absolute
value function is a good example of this: y = |x| (the graph looks
like a V). It is continuous but it has no slope at the origin, at the
corner of the V. Differentiable means it must have a derivative, a
slope, at every point.

When a graph has a derivative at a point, it is sometimes called
"locally linear." That's because if you zoom in very close to the
point, the graph appears to be a straight line. Similarly, the Earth
looks flat to us, living on its surface. But from space it appears
curved. A graph like y = x^2 will appear to be a straight line if you
look at it under a microscope. Thus, it is differentiable. The
absolute value graph, however, will always look like a V, no matter
how much you zoom in towards the vertex (0,0). You will never
straighten the graph out. So at x = 0, y = |x| is NOT locally linear
and, therefore, does not have a derivative.

The reverse proposition, by the way, is true. That is, if a function
is differentiable it MUST be continuous. If, when you zoom in on the
graph it appears to be a straight line, then it is continuous, since
straight lines are continuous.

Now on to the rest of your question:

f(x) = x^2 + 2, for x less than or equal to 1
= a(x - (1/x)) + b, for x greater than 1

f(x) is an example of a "piecewise function." That is, it is made up
of two pieces. One piece is a part of the parabola y = x^2 + 2. The
other piece is part of a weird curve, a(x - (1/x)) + b.

Depending upon the choice of "a" and "b," the two pieces need not join
together, in which case the curve will not be continuous and,
therefore, not differentiable.

On the other hand, there may be choices of "a" and "b" that will let
the pieces join together -- smoothly, or, like the absolute value
function, the pieces may form a corner, in which case it will not be
differentiable, since the function won't be locally linear at the
point of joining. That is what you are seeking: smooth curves are
differentiable.

So let's take the first part first. What values of "a" and "b" will
make this function continuous? We must find values that will let the
two pieces of the graph connect.

Recall the function:

f(x) = x^2 + 2, for x less than or equal to 1
= a(x - (1/x)) + b, for x greater than 1

The key point here is the value x = 1. That is where f(x) breaks in
two pieces. The first part of the definition shows that f(1) = 3, so
the first piece of the graph ends at the point (1, 3). If the function
is continuous, then the second piece of the graph must start at (1,3).
That is, when x = 1, we need

a(1 - (1/1)) + b = 3

Thus, b = 3.

So if b = 3, the two pieces will connect. That is true for ANY value
of "a." If you have a graphing calculator or graphing software, you
might try graphing y = x^2 + 2 and y = a(x - (1/x)) + 3 for several
different values of "a." In every case, the graphs intersect at (1,3),
but at different angles.

Now we are ready for the last part of the question. What value of "a"
will make the function differentiable? That means we want to avoid a
corner at (1,3). To do this, we need to find the derivative of each
part of the function:

When x < 1, f'(x) = 2x.
At x = 1, this piece has slope 2.

When x > 1, f'(x) = a(1 + 1/x^2).
At x = 1, this piece has slope a(1 + 1) = 2a.

You see, different choices of "a" will give different slopes for this
piece. To make f(x) differentiable at (1,3), we want the two pieces to
have the same slope. This gives us an equation:

2 = 2a

So choosing a = 1 and b = 3 will produce a function that is both
continuous and differentiable. Choosing any other value for "a" with
b = 3 will give a function that is continuous but not differentiable
at x = 1. Finally, any other choice of "a" and "b" will produce a
function which is discontinuous at x = 1 and, therefore, not
differentiable there either.

I hope that helps.

-Doctor Sam, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus
High School Functions

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search