IVT/Rolles theorem...proving a root exists

Question

Hi, I just need to know if my explanation is ok. I think the question might be on my test tomorow. It asks to prove that f(x)= x^3+3x^2+6x has one real root. My explanation: 

 

f(1)=10>0 

f(-1)= -4<0 

 

By IVT, there is a number c in the interval ( -1,1) such that f(c)=0 

 

 

Assume there are 2 roots a and b , then f(a)=f(b)=0 . So by Rolles theorem, there is a number r in the interval ( a,b) and f'(r)=0. But f'(x)= 3x^2+6x+6>0 , so you cant get f'(r)=0

Answer 1

f(x) = x³ + 3x² + 6x

 

Since f(x) is a polynomial function, it is continuous

 

f(1) = 10 > 0

f(-1) = -4 < 0

 

So, by the Intermediate Value Theorem, f(x) has at least one real root, c, between x = -1 and x = 1.

 

Assume that f(x) has another real root, b.

 

So, f(b) = 0 and f(c) = 0. f(x) is a continuous function on the closed interval with endpoints b and c and is differentiable on the open interval with endpoints b and c.

 

Therefore, by Rolle's Theorem, There is at least one real number, d, between b and c where f'(d) = 0.

 

But, f'(x) = 3x² +6x + 6 has no real roots (check by the quadratic formula).

 

**CONTRADICTION**

 

So, f(x) has only one real root.

 

Answer 2

I'm not convinced because I think that there are two solutions for f'(x) = 0; use the quadratic formula.  (decimally, these are -.42265 and -1.57735)

I withdraw this observation because I see that my f' was erroneously written as 3x squared plus 6x plus TWO.