I'm stuck on this problem. Let a and b be fixed non-zero real numbers and let f(x) = 2(b^2)x^3 + abx^2 - [(a^2) + (3b^2) + 1] - 3ab. Prove that f(x) = 0 has three real roots and one of these roots must lie in the interval [-1,1]. Thank you for your help.
I'm pretty sure I know how to prove that one root lies in the interval [-1,1]. I find f(-1) and f(1) and then use the Intermediate Value Theorem since f(-1) is less than zero and f(1) is greater than zero. However, how do I show that there are three roots? Can someone just show me this?
By the way, there should be an x after the brackets in my original function f(x).
You can try to graph the function to see if it intersects x-axis three times. But since we do not know exactly the values of a,b it is not that easy to do. However, we can approximate how it looks like. For example, find f'(x) and set it as zero so x=-1/3. That is a relative max/min which is inside the interval. Now try the argument that says it is negative or something like that.
The b, a, and c refers to:
If Delta is:
1. Greater than zero and a perfect square, then the roots are Real, Rational, and not equal.
2. Greater than zero but not a perfect square, then the roots are Real, Irrational, and not equal.
3. Equal to zero. The roots are Real, Rational, and Equal.
4. Smaller than zero. The roots do not exist.