Use a cosine amplitude bound that works for every phase and frequency

$y=2x+2+3/(x-1)$, with asymptotes x=1 and y=2 x+2. There is no x-intercept since 2 x²+1>0; the y-intercept is (0,−1). The turns are $(1-\sqrt6/2,4-2\sqrt6)$, maximum≈(−0.225,−0.899), and $(1+\sqrt6/2,4+2\sqrt6)$, minimum≈(2.22,8.90), retaining the source labels.

Dividing the equation by x−1 gives $(2x^2+1)/(x-1)=a\cos(bx+c)$. The source instructs sketching the cosine on the rational graph. For any b,c, its outputs lie in[−|a|,|a|]. The rational curve has outputs y≤4−2√6 or y≥4+2√6. Thus the symmetric cosine band avoids it if $|a|<2\sqrt6-4$.

The exact range is $4-2\sqrt6<a<2\sqrt6-4$. The source writes −0.899<a<0.899 using the rounded maximum magnitude; these are approximate boundary labels, not exact endpoints. At |a|=2√6−4 one can choose a phase giving the rational curve’s maximum; for larger |a|one can also choose b=0 and a phase attaining an available negative output. Therefore the strict bound is necessary for the guarantee for all b,c.