FunctionHard
Question
Let f(x) = (1 + b2)x2 + 2bx + 1 and let m (b) the minimum value of f(x) As b varies, the renge of m (b) is
Options
A.[0, 1]
B.
C.
D.(0,1]
Solution
Given, f(x) = (1 + b2)x2 + 2bx + 1
= (1 + b2)
m (b) = minimum value of f(x)
is positive and
m (b) varies from 1 to 0, so range = (0,1]
= (1 + b2)
m (b) = minimum value of f(x)
is positive and m (b) varies from 1 to 0, so range = (0,1]
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