<s>[INST] Maxima and minima of $\frac{x^2-3x+4}{x^2+3x+4}$ without calculus How to find the minimum and max values of $y=\frac{x^2-3x+4}{x^2+3x+4}$ for all real values of $x$ without using calculus?
Perhaps it could be done graphically by noting the fact that the numerator and denominator are a pair of parabolas symmetric about the $x$ axis, but I do not know how to continue.
Thanks!
 [/INST] Let the maximum of $f(x) = \frac{x^2-3x+4}{x^2+3x+4}$ be $m$. Then:
$$x^2-3x+4 = mx^2 + 3mx + 4m$$
$$(m-1)x^2 + (3m+3)x + (4m - 4) = 0$$
We want this equation to have only one real root (a double root), so:
$$\Delta = 0 \Rightarrow (3m+3)^2-4(m-1)(4m-4) = 0.$$
A similar process for the minimum ($n$) yields the same equation, as multiplying it by $-1$ does not change the values of $m$. Therefore, both the maximum and minimum values are given by this equation.
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