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### Differentiability, Intervals, Inflection Points of Piecewise Function

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Date: 05/13/98 at 19:47:26
From: Cathy Emineth
Subject: AP Calculus: Integral

Okay, this is my problem (from an AP exam):

Let f be a function defined by f(x) = {2x - x^2 for x < or = 1
{x^2 + kx + p for x > 1

(a) For what values of k and p will f be continuous and
differentiable?

(b) For the values of k and p found in (a), on what interval or
intervals is f increasing?

(c) Using the values of k and p found in part (a), find all points
of inflection of the graph of f. Support your conclusion.

Yuck, right? This is an AP essay question, which means that all work
MUST be shown and clearly labeled.

For starters, I set x^2 + kx + p = 1. I found that this worked when
k = 1, p = -1, or vice versa. Then all values greater than one work in
this equation.

For part (b), I drew a blank. Help!

This is pre-1992, which means no calculators of any sort are allowed.

My class has been getting one of these lovelies every week since the
beginning of the year. I consider this unfair, because I never planned
on taking the AP exam, but AP Calculus is the only course offered in
my school for the senior year besides Senior Math.
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Date: 05/16/98 at 09:07:13
From: Doctor Schwa
Subject: Re: AP Calculus: Integral

Thanks for letting me know how you started working on the problem.
That helps me understand what you're thinking about and lets me write
an answer that (I hope!) is more useful to you.

For a function to be continuous, as you point out, x^2 + kx + p must
equal 1. But at what value of x? The transition between the two
"pieces" of the function happens at x = 1. So that's where the values
must match. You correctly plugged in x = 1 to find 2x - x^2 = 1, but
you also have to plug in x = 1 at the other end, so that the limit as
x->1 from below equals the limit as x->1 from above, in order for the
function to be continuous.

The function also has to be differentiable, so the slope from below
must equal the slope from above. That is, the derivative 2 - 2x, which
approaches 0 as x->1, must equal 2x + k, which approaches 2 + k
as x->1.

That's enough information to solve for k and p, and there is only one
possibility. (You have two facts, that the function must match up and
the slope must match up; and you have two variables, k and p, which is
usually enough information to solve the problem completely).

as much calculus is to think about (2x - x^2) as a downward-opening
parabola that has just hit its maximum (vertex) at x = 1. So to
continue it with the upward-opening parabola x^2 + kx + p, we have to
have its vertex also at the same point (1,1), which gives you another
way to find values for k and p.

Now we could find where it's increasing by taking derivatives:

2 - 2x when x < 1
2x - 2 when x > 1

and then check where the first derivative is positive. You might
notice at x = 1 that the derivative is zero. Does that mean it's not
increasing there? It depends on how you define "increasing." There was
let me know if you're interested in talking more about the various
possible ways of defining an increasing function.

You could also use the graph of these two half-parabolas to figure out
when it's increasing (and again, the vertex point (1,1) will be a
special case, depending on your definition of "increasing").

For part (c), you can take the second derivative and see if there are
any points where it changes sign. Those would be inflection points. I
find it convenient to draw a number line to keep track of the
intervals where the second derivative is positive or negative.

It is a shame that there aren't other choices of courses, such as the
new AP Statistics. What is Senior Math like? Is the problem that
there's nothing of intermediate level between that and the AP calc? At
my school, Gunn HS in Palo Alto, CA, we have AP Stat, Calc AB, and
Calc BC, as well as Precalculus, which is probably pretty similar to

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

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