The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

The Critical Number

Date: 03/23/98 at 20:17:30
From: Jen MacDnald
Subject: The critical numbers of a sine cosine function

I need to find the critical number of the following equation:

    y  = 2 cos x + sin 2x  

where x is greater or equal to 0 and less than or equal to 2 Pi.

In order to find critical numbers you need to take the derivative, 
which I did, and got:

    y' = [(0)(cos x)+2(-sin x)] - cos 2x (0)(x)

which reduced equals:

    y' = -2 sin x - x cos 2x

I then set y' to zero and got:

     0 = -2 sin x - x cos 2x

This is as far as I got because I don't know how to manipulate the 
equation to find out what x equals.  I see that I can possibly use 
the double angle formula to solve for x but I don't know which one 
or if it will even help at all.

Date: 03/24/98 at 11:17:05
From: Doctor Sam
Subject: Re: The critical numbers of a sine cosine function


Part of your problem is that you don't have the correct derivative.  
If y = 2 cos x + sin 2x  then y' = -2sin x + 2cos 2x.  You went wrong 
with the Chain Rule in the second function.

    [sin 2x]' = cos 2x [2x]' = 2 cos 2x    

I don't know where you got the "x" from or the "0".  If there were 
a factor of zero then instead of getting x cos 2x you should have 
gotten zero for that term.

I suspect that you differentiated the "2" to get zero. You need to 
practice how to handle constants when you are differentiating. It is 
true that the derivative of 2 is zero, but this 2 never gets 
differentiated by itself.

I think about it this way:  [2x]' = 2 because the slope of the line 
y = 2x is 2. If you are not convinced but have a graphics calculator 
available, try graphing y = cos 2x and NDeriv(cos 2x, x, x) [which is 
how the Texas Instruments calculator approximates a derivative].
You should see the graph of y = cos 2x and the graph of - 2 sin 2x.

Before I get to your "critical points" question, I have one more 
comment on your differentiation. You used the product rule on 2sin x.  
You used the rule correctly and got the right answer, but you didn't 
need the product rule.  

    [2 sin x]' = 2[sin x]' = 2 cos x

In general, coefficients just get carried along as multipliers when 
you take the derivative. That's why [3x^2]' = 6x; you multiply the 2 
and the 3 to get the new coefficient because 

       [3x^2]' = 3[x^2]' = 3[2x] = 6x 

I think that you are likely to make more computational errors if you 
use a complicated rule like the product rule for simple expressions 
like 2sin x or the quotient rule for (sin x)/2.

Okay, now to your question. If 

            y  = 2cos x+ sin 2x,      then 
            y' = -2sin x + 2cos 2x.

Set the derivative equal to zero: 0 = -2sin x + 2cos 2x.

This is hard to solve because one of the trig functions is in terms 
of x and the other in terms of 2x.  This is a good place to use the 
double angle formula for cos 2x, which lets us rewrite a '2x' trig 
function in terms of just 'x' trig functions.  

There are several ways of doing this. Since there is a sin x in the 
equation I'm going to use 

       cos 2x = 1 - 2 (sin x)^2.  This gives:

            0 = -2sin x + 2[ 1 - 2(sin x)^2] 
            0 = -2sin x + 2 - 4(sin x)^2

This is complicated but it is really just a quadratic equation. It may 
help to substitute z = sin x to see this:

            0 = -2z + 2- 4z^2  

Divide through by -2 and rearrange to get:

            0 = 2z^2 + z - 1     To solve this, factor:
            0 = (2z - 1)(z + 1)  and put back the sin x:
            0 = (2sin x - 1)(sin x + 1)

A product is zero only when one or more of its factors are zero.  
So either 
  2 sin x - 1 = 0    or 
    sin x + 1 = 0

If 2sin x - 1 = 0 then sin x = 1/2.  There are two places between 
0 and 2pi where sin x = 1/2.  In the first quadrant x = pi/6  and 
in the second quadrant x = 5pi/6.

If sin x + 1 = 0 then sin x = -1. There is only one place between 
0 and 2pi where sin x = -1, at x = 3pi/2.

So there are three places where y' = 0:  x = pi/6, 5pi/6, and 3pi/2.

There may be two more values depending upon the question. If you just 
want to know where y' = 0 between 0 and 2pi then we have all three 
values. But if the problem is to find the critical values of 
y = 2cos x - sin 2x on the domain 0 <= x <= 2pi then the problem is 
a bit different.  

In the first case, the domain of y = 2cos x - sin 2x is all real 
numbers and you are only looking for places where the graph has a 
horizontal tangent.

In the second case, the domain is a closed interval. Whenever a 
function is defined on a closed interval, its endpoints are also 
critical values. The reason for this is that what makes a point 
critical is that it may be a location of a maximum y-value or 
a minimum y-value. For functions whose domain is all real numbers, 
high points and low points can only occur at places where the 
derivative is zero (like x = 0 for y = x^2) or where the derivative 
doesn't exist (like x = 0 for y = |x| which is also a minimum).

But when a graph has a closed domain like 0 <= x <= 2pi the endpoints 
of the graph are also locally high or low points. For example, y = 3x 
is just a line; it has no maximum or minimum. But if we only look at 
y = 2x for x between 2 and 4 then we have just a line segment. The 
point (2, 6) is the lowest point on the segment and the point (4, 12) 
is the highest point, so both x = 2 and x = 4 are critical values, 
even though the derivative of y = 3x is y' = 3, which is never zero.

So if the problem is to find the critical values of 
y = 2cos x - sin 2x on the closed interval 0 <= x <= 2pi, then there 
are two more critical values: x = 0 and x = 2pi.

I hope that helps.

-Doctor Sam,  The Math Forum
Check out our Web site   
Associated Topics:
High School Calculus

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- The Math Forum at NCTM. All rights reserved.