Intermediate value theorem

[iRubeeeeen]

Hello. I'm studying for my analysis final and I have no idea how to do this problem. It says the following:

 

Suppose f(x)=x^3+3x^2-2x-1 is continuous everywhere on R. Use the intermediate value theorem to show that f(x) has three distinct real roots. Hint: Evaluate f(x) at x=-1,0, an 1 and make use of the limit behavior of f(x) as x approaches plus or minus infinity.

 

I calculated the values and got that f(-1)=3, f(0)=-1, and f(1)=1. Also, as x approaches negative infinity, f(x) approaches negative infinity and as x approaches positive infinity, f(x) approaches positive infinity. When I graph it, I can see that there are three real roots but I just don't know how to go about the proof. Any suggestions or help? Thank you.

[Acme]

Hello. I'm studying for my analysis final and I have no idea how to do this problem. It says the following:

 

Suppose f(x)=x^3+3x^2-2x-1 is continuous everywhere on R. Use the intermediate value theorem to show that f(x) has three distinct real roots. Hint: Evaluate f(x) at x=-1,0, an 1 and make use of the limit behavior of f(x) as x approaches plus or minus infinity.

 

I calculated the values and got that f(-1)=3, f(0)=-1, and f(1)=1. Also, as x approaches negative infinity, f(x) approaches negative infinity and as x approaches positive infinity, f(x) approaches positive infinity. When I graph it, I can see that there are three real roots but I just don't know how to go about the proof. Any suggestions or help? Thank you.

I had to read up on the intermediate theorem @ Wiki and based on that reading I will take an amateur stab. As the function is given as continuous and you have already established an interval between negative and positive infinity, then by the intermediate value theorem

...If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano's theorem).[1] And, the image of a continuous function over an interval is itself an interval.

So the way I read things, that's the proof that [at least] 3 roots exist.

 

If that is circular or otherwise a wild swing and you or someone you know has been injured, I was never here.

Edited June 8, 2014 by Acme

[imatfaal]

Another amateur hour attempt from me!

 

You have four intervals

1. neg infin --> -1

2. -1--> 0

3. 0 -->1

4. 1--> pos infin

 

1. f(x) starts neg and ends positive

2. f(x) starts pos and end neg

3. f(x) starts neg and end pos

4. f(x) starts post and ends pos

 

 

The part that Acme quoted allows you to state that in intervals 1. 2. & 3 there must exist at least one root (or any odd number of roots). And as a cubic can have at most three roots then there must be only one in each interval. Note interval 4. could have two roots (or any even number) if it were not for maximum number of roots of a cubic

 

 

f(-1)=3, f(0)=-1, and f(1)=1

[studiot]

The general cubic y = x³ + ax² + bx + c is shown below for various values of coefficients a, b & c.

 

 

For values of x that are very large and positive [math]y \mapsto \left( { + \infty } \right)[/math]

For values of x that are very large and negative [math]y \mapsto \left( { - \infty } \right)[/math]

 

So it must cross the x axis at least once on its continuous journey from negative infinity to positive infinity.

imatfaal has pointed out that a cubic has three roots so if it crosses the x axis a second time ie left to right or from x negative to x positive it must cross back again to y positive.

 

So a general cubic has one or three real roots.

 

By evaluating the signs at the hinted points you can establish whether there is one or three crossings and where they are, following the same logic within each interval as others have indicated.

Should this not be in homework help?

Edited June 8, 2014 by studiot