Question

if f(x)=(1)/(ax^2+bx+4) and is a even function what is the value of b?

Answer 1

This means that (1)/(ax^2 + bx + 4) = (1)/[a(-x)^2 + b(-x) + 4]

ax^2 will always equal a(-x)^2, so we have to find a "b" where bx = b(-x) and that will only happen if:

b = 0

Answer 2

but if it is 0 it is 1/x^2+4 = 1/(x-2)(x+2) which isnt even

Answer 3

x^2 + 4 does not = (x - 2)(x + 2)

x^2 - 4 does equal that though but that is not relevant here.

Answer 4

What is relevant is that 1/(x^2 + 4) = 1/[(-x)^2 + 4] for all "x"

Answer 5

ah i see you are correct... much obliged

Answer 6

you're welcome!