Algebra> let f(x) = (1+b^2)x^2 + 2bx + 1 and m(b) ...

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

Gaurang Kotasthane , 8 Years ago

Grade 11

3 Answers

Sarita Sharma

Last Activity: 8 Years ago

Since this is a quadratic equation its minimum is -D/4a (by differentiation) D=b^2 -4ac m(b)

Ajay

Last Activity: 8 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

Last Activity: 8 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