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$$f(x) = \frac{\left(\cos x\space +\space 0.5\right)}{\left(1\space +\space 0.5\cos x\right)^2}$$

Looking at the graph I know that the function is never undefined, but how would I show this or explain it algebraically?

Also how come when I change 0.5 (in both the numerator and denominator) to some number greater than 1, it becomes undefined?

Does it have something to do with cosx and the fact that its periodic?

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  • $\begingroup$ Can the denominator vanish for some $x\in\mathbb{R}$? $\endgroup$ – Ángel Mario Gallegos Jan 10 '15 at 5:12
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For this to be undefined, the denominator has to be zero. For the denom to be $0$ we must have,

$1=-0.5\cos(x)$, which gives $\cos(x)=-2$. But $|\cos(x)| \leq 1$ for all $x$, and hence the denom can never be $0$.

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In the function $$f(x) = \frac{\cos (x) + 0.5}{\left(1 + a\cos x\right)^2}$$ the numerator does not make any problem. For the denominator, it is another story since it could cancel for some $x$ such that $\cos(x)=-\frac 1a$; this cannot happen if $a<1$ since $-1\leq \cos(x)\leq 1$. But, if $|a|>1$, the denominator cancels and this is the plave where the function is undefined.

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