Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
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


Let's start with your last question first. The answer is no. A 
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 

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

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994-2013 The Math Forum