Find the maximum and the minimum value of x if $\frac{\sqrt{100-x^2}+\sqrt{99+x^2}}{40}= \cos \frac{\pi}{x^2-2|x|+4}$.

Question

Find the maximum and the minimum value of $x$ if $$\frac{\sqrt{100-x^2}+\sqrt{99+x^2}}{40}= \cos \frac{\pi}{x^2-2|x|+4}$$

I just did $100-x^2\geq0\implies-10\leq x\leq10$, but what I have to do now? Is the answer -10 and 10?

Answer 1

Since $$x^2-2|x|+4\geq 3$$ we have $${\pi\over 3} \geq {\pi\over x^2-2|x|+4}>0$$

so $${1\over 2}= \cos {\pi\over 3} \leq \cos {\pi\over x^2-2|x|+4}$$

so $$\sqrt{100-x^2}+\sqrt{99+x^2}\geq 20$$ and this probably has no solution (just calculate the minimum of the left side, say with the derivative). So there is no such $x$.

Edit: Since $$ \sqrt{2(a^2+b^2)} \geq a+b$$ we see that left side is at most $\sqrt{398}<20$ so there is no solution.

Answer 2

$$\frac{\sqrt{100-x^2}+\sqrt{99+x^2}}{40}= \cos \frac{\pi}{x^2-2|x|+4}$$ $\cos()$ is a periodic function and it's minimum value is $-1$ it's maximum value is $1$

Note that your statement is not an inequality, but some equation, the maximum and minimum value of $x$ are roots of the equation, solving in by Newton approximation might not be helpful enough so you can plot the function to check or use the method of derivatives