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let f(x) = (1+b^2)x^2 + 2bx + 1 and m(b) be the minimum of f(x) as b varies the range of m(b) is

let f(x) = (1+b^2)x^2 + 2bx + 1 and m(b) be the minimum of f(x) as b varies the range of m(b) is 

Grade:11

3 Answers

Sarita Sharma
9 Points
7 years ago
Since this is a quadratic equation its minimum is -D/4a (by differentiation) D=b^2 -4ac m(b)
Ajay
209 Points
7 years ago
Nice question 
As Sarita mentioned the minimum value of F(x) is  -D/4a
Here D = 4b2-4(1+b2) = -4 and
4a =  4 (1+ b2)
Hence Minimum value is  1/(1+b2) = m(b)
 Now  the given question is to find range of m(b) whicj can be found as follows
let y =  1/(1+b2)
b = sqrt ((1-y)/y)
 
Since b is real follwing conditions should be satisfied
(1-y)/y greater or equal to zero and y not equal to zero.
 
Solving the inequality for y the range of b is (0 1]
 
 
mycroft holmes
272 Points
7 years ago
You can write f(x) = x2+(bx+1)2
 
Consider the line y = bx+1. Any point P on this line is given by (x,bx+1)
 
So f(x) = OP2 is minimum when P is the foot of the perpendicular from O (origin) on to the line i.e. we are looking at the distance from O to the line. This as we know from the formula is simply, 1/(b2+1). The max va;ue for the expression is easily seen to be 1 when b=0

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