Proving Existence of Real Solutions for Polynomial Equations

In summary, the equation f(x) = 0 has at least one real solution because the function f is continuous and the intermediate value theorem applies. By showing that f(-1) = 4 and f(-2) = -504, it is proven that there must be some real number c between -2 and -1 such that f(c) = 0. A formal proof would involve stating the continuity of f, finding f(-2) and f(-1), and using the IVT to show the existence of a real solution.
  • #1
Janez25
19
0

Homework Statement


Let f(x) = x^9+x^2+4. Prove: The equation f(x)=0 has at least one real solution.


Homework Equations





The Attempt at a Solution


I know that the solution lies between -2 and -1. I also know that f(-2) = -504 and f(-1) = 4. I need to know how to use the IVT to prove that there is one real solution; not sure how to do that.
 
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  • #2
You've already done it. f is continuous so the intermediate value theorem applies. You showed that f(-1) = 4 and f(-2) is -504, and therefore there must be some c between -2 and -1 for which f(c) = 0 since 0 lies between 4 and -504.
 
  • #3
How would I write this up as a formal proof?
 
  • #4
First you point out that f is continuous, and say that the IVT applies. Then you find f(-2) and f(-1), and say that 0 lies between these two values because one is positive and one is negative. Then, by the IVT, there must be some real number c between -2 and -1 such that f(c) = 0.
 
  • #5
Thanks so much for all your help!
 

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