Assassin315
New member
- Joined
- Oct 26, 2010
- Messages
- 12
A quick problem here. I'm in a calculus class where we've got to relearn some basic trig, so let's see how this goes. I'll provide my thoughts/answers to each question to see if my process is sound.
The problem provides the equation "f(x) = anx^n + an-1x^n-1 + ... + a1x^1 + a0," and it assumes n is any odd number.
***Part 1 asks to prove that, if an > 0, then the limit of f(x) as x goes to negative infinity is negative infinity, and the limit of f(x) as x goes to positive infinity is positive infinity.
Now, I'm not entirely sure how to formally prove this. I know that the anx^n term controls the equation's end behavior. If you let x go to infinity, regardless of the other terms, f(x) goes to infinity. Likewise, since n is any odd number, letting x go to negative infinity will result in f(x) going to negative infinity. Do you think there's a better way to explain this? I've never been that good with proofs.
***Part 2 says that, using part 1, explain that there are numbers a, b such that f(a) < 0 and f(b) > 0.
I feel like I'm really oversimplifying my answer for this. Part 1 basically proved that the range for the given polynomial f(x) is (-inf.,inf.), so there *must* exist some points a and b where the values are either less than or greater than 0. Too easy?
***Part 3 asks what happens when a < 0.
Since we already analyzed the end behavior in part 1, I think this is also pretty straightforward: the limits are switched, right? If a < 0, then x going to infinity results in a negative number; x going to negative infinity results in a positive number.
So mostly, I'm not sure about the first part, but I'm mostly sure about the last two. Thoughts? Corrections?
The problem provides the equation "f(x) = anx^n + an-1x^n-1 + ... + a1x^1 + a0," and it assumes n is any odd number.
***Part 1 asks to prove that, if an > 0, then the limit of f(x) as x goes to negative infinity is negative infinity, and the limit of f(x) as x goes to positive infinity is positive infinity.
Now, I'm not entirely sure how to formally prove this. I know that the anx^n term controls the equation's end behavior. If you let x go to infinity, regardless of the other terms, f(x) goes to infinity. Likewise, since n is any odd number, letting x go to negative infinity will result in f(x) going to negative infinity. Do you think there's a better way to explain this? I've never been that good with proofs.
***Part 2 says that, using part 1, explain that there are numbers a, b such that f(a) < 0 and f(b) > 0.
I feel like I'm really oversimplifying my answer for this. Part 1 basically proved that the range for the given polynomial f(x) is (-inf.,inf.), so there *must* exist some points a and b where the values are either less than or greater than 0. Too easy?
***Part 3 asks what happens when a < 0.
Since we already analyzed the end behavior in part 1, I think this is also pretty straightforward: the limits are switched, right? If a < 0, then x going to infinity results in a negative number; x going to negative infinity results in a positive number.
So mostly, I'm not sure about the first part, but I'm mostly sure about the last two. Thoughts? Corrections?