Expert: Paul Klarreich - 2/3/2006

Question
Hello! Since you'd helped me immensely before with a problem, I thought I'd pester you again about this one! I must be missing something because I just can't get this one. Here's the problem:

Consider the function f(x)=x^3 + ax^2 + c. Show that if a < 0 and c > 0, then f has exactly one negative root.

In this chapter, we're studying Rolle's theorem, and the Mean-Value Theorem. So I assumed they want us to use these to solve the problem.

First, since it's a polynomial of odd degree, you know it has at least one real root. Then, I found the derivative: f'(x)=3x^2 + 2ax, and set this equal to zero to find out if it could have any others. To prove that f has only one root, you'd expect to find f'(x)=0 has no solutions, but I got 0 and 2a/3. Then I plugged these values of x into f(x) and got y=C and y=((20a^3)/27) + c.

I tried plugging in some numbers and graphing the function to get an idea what the root might be, but when I did that I found if a < -2, the function has 3 roots. Am I missing something obvious? I don't see how this function can have only one negative root.

Thank you for any help!  

Answer
Hi, Kayla,

>> Thank you for the kind comments.

Your Question:  Hello! Since you'd helped me immensely before with a problem, I thought I'd pester you again about this one! I must be missing something because I just can't get this one. Here's the problem:

Consider the function f(x)=x^3 + ax^2 + c. Show that if a < 0 and c > 0, then f has exactly one negative root.

In this chapter, we're studying Rolle's theorem, and the Mean-Value Theorem. So I assumed they want us to use these to solve the problem.

First, since it's a polynomial of odd degree, you know it has at least one real root. Then, I found the derivative: f'(x)=3x^2 + 2ax, and set this equal to zero to find out if it could have any others.

To prove that f has only one root,

>> I thought you had to prove that f has only one NEGATIVE root.

you'd expect to find f'(x)=0 has no solutions, but I got 0 and 2a/3.

>> I think you blew a sign here.  Isn't that -2a/3?


Then I plugged these values of x into f(x) and got y=C and y=((20a^3)/27) + c.

>> function values at these points are not totally relevant.

I tried plugging in some numbers and graphing the function to get an idea what the root might be, but when I did that I found if a < -2, the function has 3 roots. Am I missing something obvious? I don't see how this function can have only one negative root.

Thank you for any help!  
---------------------
>> OK.  You made some progress, now it's time to draw some conclusions.  

Yes, it has at least one real root, since it is a cubic.
Now you found your f'(x) and set it to zero and found:

f'(0) = 0  and  f'(-2a/3) = 0  

Rolle's theorem says if x1 and x2 are zeroes of a well-behaved function f(x), (and polynomials are that) then f'(x) has a zero between x1 and x2.  

So where are your x1, x2, x3 that are roots of your f(x)?  Well, your zeroes of the derivative are:

At x = 0, so at least one of the roots must on each side of x = 0
At x = -2a/3, which is a POSITIVE NUMBER, because you said a<0.

So, you might have:

A zero to the left of x = 0, so it is negative.
A zero between x = 0 and x = -2a/3, so it is positive.
A zero to the right of x = -2a/3, so it is also positive.

So if the function has three zeroes, only one can be negative.  

Now which of these are MUST HAVE's?  We have 'turning points' at x=0 and at x=-2a/3, which is positive.  And clearly the one at x=0 is a maximum.  (Easy to prove by the second derivative test.)  

Now if this maximum is below the x-axis, the graph comes up from deep in Quadrant-3, turns before reaching the x-axis, goes down, turns up at the second turning point, then heads up into Q-1.  In that case, it will have only the one root which is positive.

But the fact that c>0 (you thought we would never get around to that, didn't you.) means that f(0) = c > 0.  So the first turning point is ABOVE the x-axis.  That means, that as the graph comes up from Q-3, it must cross the x-axis to the left of x=0, and it must have a negative root.  

Whew!