Proving Existence of Real Solutions for Polynomial Equations

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Homework Statement


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


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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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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.
 
How would I write this up as a formal proof?
 
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.
 
Thanks so much for all your help!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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