As $ b$ varies range of $m(b)$ is? Let $f(x)=(1+b^2)x^2+2bx+1$ and let $m(b)$ be the minimum value of $f(x)$.As $b$ varies range of $m(b)$ is ?
I can find the minima by putting $f'(x)=0$ and hence find $m(b)$.But after that i'm not able to find range of $m(b)$.
 A: The minima occurs at$\frac{-b}{1+b^2}$
Substituting this we get,
$$m(b)=1-\frac{b^2}{1+b^2} $$
$$m(b)=\frac{1}{1+b^2} $$
Therefore range of m(b)=(0,1]
A: Finding minimum :
$$f(x)=(1+b^2)x^2+2bx+1$$ 
$$f(x)=(1+b^2)\left[x^2+\frac{2b}{1+b^2}x+\frac{1}{1+b^2}\right]$$ 
$$f(x)=(1+b^2)\left[\left(x+\frac{b}{1+b^2}\right)^2-\frac{b^2}{(1+b^2)^2}+\frac{1}{1+b^2}\right]$$ 
$$f(x)=(1+b^2)\left[\left(x+\frac{b}{1+b^2}\right)^2-\frac{b^2}{(1+b^2)^2}+\frac{1+b^2}{(1+b^2)^2}\right]$$
$$f(x)=(1+b^2)\left[\left(x+\frac{b}{1+b^2}\right)^2+\frac{b^2}{(1+b^2)^2}\right]$$
Thus $$f(x)=(1+b^2)\left(x+\frac{b}{1+b^2}\right)^2+\frac{1}{1+b^2}$$
Since $$\left(x+\frac{b}{1+b^2}\right)^2 \geq 0$$
$$(1+b^2)\left(x+\frac{b}{1+b^2}\right)^2 \geq 0$$
$$(1+b^2)\left(x+\frac{b}{1+b^2}\right)^2+\frac{1}{1+b^2} \geq \frac{1}{1+b^2}$$
$$f(x) \geq \frac{1}{1+b^2}$$
Therefore $$f(x)_{min}=m(b)=\frac{1}{1+b^2}$$
Finding the range of the minimum :
Now take $y=m(b)$ , for all real $b$ ,
$$y=\frac{1}{1+b^2} \Rightarrow y+yb^2=1$$
$$yb^2+y-1=0$$
Since $b$ is a variable , this is a quadratic equation of $b$. 
Thus for all real $b$ , its discriminant must be greater than or equal to zero.
That is , $$0^2-4y(y-1) \geq 0$$
$$(y-1)y \leq 0$$
Thus $$0 < y \leq 1$$ 
