Let $\text{⨍ (x)=}\left(1+{\mathrm{b}}^2\right){\text{x}}^2\text{+2bx+1}$ 1 and let m(b) be the minimum value of $\text{⨍ (x).}$ As b varies, the range of m (b) is.

$\begin{array}{l}f\left(x\right)=(1+b^2)x^2+2bx+1\\ m\left(b\right)=\min imumvalue=\frac{4ac-b^2}{4a}\\ =\frac{4(1+b^2)-4b^2}{4(1+b^2)}\\ =\frac{1}{1+b^2}\\ Lety=\frac{1}{1+b^2}\\ \Rightarrow y+y{b}^2=1\\ \Rightarrow y{b}^2=1-y\\ \Rightarrow b=\sqrt{\frac{1-y}{y}}isdefined\\ If\frac{1-y}{y}\ge 0andy\ne 0\\ \therefore y(1-y)\ge 0\\ \Rightarrow y(y-1)\le 0\\ \Rightarrow 0<y\le 1\end{array}$