Solve x3-6x2+1=0: Find Real Numbers & Reasons

[Shadowless]

Homework Statement

Determine how many real numbers satisfy the equation x³-6x²+1=0. Give reasons for your answer naming any theorems you use.

2. The attempt at a solution

let:f(x)=x³-6x²+1

→f'(x)=3x²-12x

for stationary points:f'(x)=0

→3x(x-4)=0

∴x=0 or x=4

to determine if minimum or maximum points:

f''(x)=6x-12

→f''(0)=-12<0

→f''(4)=12>0

∴(0,1) is a maximum point.

∴(4,-31) is a minimum point.

Not sure how to finish. I know it has 3 solutions, but I'm not sure as how to express my answer in the correct format. Help?

 

[SammyS]

Homework Statement

Determine how many real numbers satisfy the equation x³-6x²+1=0. Give reasons for your answer naming any theorems you use.

2. The attempt at a solution

let:f(x)=x³-6x²+1

→f'(x)=3x²-12x

for stationary points:f'(x)=0

→3x(x-4)=0

∴x=0 or x=4

to determine if minimum or maximum points:

f''(x)=6x-12

→f''(0)=-12<0

→f''(4)=12>0

∴(0,1) is a maximum point.

∴(4,-31) is a minimum point.

Not sure how to finish. I know it has 3 solutions, but I'm not sure as how to express my answer in the correct format. Help?

Note: The above quote has been edited to reflect changes made by Shadowless after I initially posted the following. The following comments stand as originally submitted by me.

Hello Shadowless. Welcome to PF !


On what interval(s) is f'(x) positive?

On what interval(s) is f'(x) negative?

 

[Shadowless]

Glad to be here.

Hmm..

f'(x) is positive for x<0 or x>4
f'(x) is negative for 0<x<4

Yes?

 

[SammyS]

Glad to be here.

Hmm..

f'(x) is positive for x<0 or x>4
f'(x) is negative for 0<x<4

Yes?

That should tell you something about the maximum number of roots that can exist in each interval: (-∞, 0), (0, 4), and (4, +∞) .

In addition to that, knowing f(0) and f(4) tells you something very definite about the number of roots between x=0 and x=4 .

Knowing the behavior of f as x → ±∞ will insure that you can find other places to evaluate f to definitely determine the number of roots.



Of course, we know that f(x) is continuous for x being any real number, right?

 

[Shadowless]

Can I just answer this question like this...

Limiting x to -∞, f(x) will approach -∞.

And as f(0)=1 and f'(x)>0 for (-∞,0) there exists only one root in the interval (-∞,0). [IVT]

As f(0)=1, f(4)=-31 and f'(x)<0 for (0,4) there exists only one root in the interval (0,4). [IVT]

Limiting x to ∞, f(x) will approach ∞.

And as f(4)=-31 and f'(x)>0 for (4,∞) there exists only one root in the interval (4,∞). [IVT]

∴ there exists three real numbers that will satisfy the equation.

 

[Dick]

Can I just answer this question like this...

Limiting x to -∞, f(x) will approach -∞.

And as f(0)=1 and f'(x)>0 for (-∞,0) there exists only one root in the interval (-∞,0). [IVT]

As f(0)=1, f(4)=-31 and f'(x)<0 for (0,4) there exists only one root in the interval (0,4). [IVT]

Limiting x to ∞, f(x) will approach ∞.

And as f(4)=-31 and f'(x)>0 for (4,∞) there exists only one root in the interval (4,∞). [IVT]

∴ there exists three real numbers that will satisfy the equation.

Yes, exactly. There will be one root in each of those intervals.

 

[Shadowless]

But would my answer be acceptable in the exam? Or would I need something further? I think I understand the question, it's just I sometimes fail to get my ideas across on paper and that is what worries me more.

 

[Dick]

But would my answer be acceptable in the exam? Or would I need something further? I think I understand the question, it's just I sometimes fail to get my ideas across on paper and that is what worries me more.

I wouldn't worry about it. Your answer is as complete and clear as anyone could wish for. You might want to mention that f(x) is continuous so you can apply the IVT, but that's all I can think of. If you want to go nuts then you could also say that the MVT proves there is only one root in each interval, but I think what you have is fine.