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?
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1I’m voting to close this question because this is not a "do-my-homework" service. – uniquesolution Apr 21 '20 at 20:15
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You found the set of all $x$ for which the left side is defined. But not all of those $x$ satisfy the equation, right? The right side must also be defined (no division by zero), and the two must be equal. – MPW Apr 21 '20 at 20:21
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$x^2-2|x|+4$ is always positive – Tas Apr 21 '20 at 20:22
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@uniquesolution Why did you think this? I just want a good person to told me the steps for solve this. I have never seen a question like this before. I just want more knowledge. – Tas Apr 21 '20 at 20:25
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@Tas first try finding out the range of $x^2 - 2|x| + 4$ and try to conclude the values RHS can lie in. Now try finding out if LHS can attain those values or not. You will find an interesting relation. – sudeep5221 Apr 21 '20 at 20:31
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The first question you should consider is whether this equation has any solutions at all. – Allawonder Apr 21 '20 at 21:04
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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.
User2020201
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1No derivaives needed. Squaring the left hand side give $199+2\sqrt{(100-x^2)(99+x^2)}<199+2\sqrt{(100-x^2)(100+x^2)}=199+2\sqrt{100^2-x^4}\le 399<20^2$ – Hagen von Eitzen Apr 21 '20 at 20:38
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$$\frac{\sqrt{100-x^2}+\sqrt{99+x^2}}{40}= \cos \frac{\pi}{x^2-2|x|+4}$$ $\cos()$ 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