Question

If f(x) =x²+kx+1,for all x and if it is an even function, find k.

Answer 1

Solution:

By definition, a function f(x) is even if:
f(x) = f(−x).

Given that: f(x)=x^2+kx+1

or, f(−x)=(−x)^2 +k(−x) + 1

or, f(-x) =(−x)(−x) − kx + 1

or, f(-x) =x^2 − kx + 1

We know that by definition, a function f(x) is even if: f(x)=f(−x)

or, x^2+kx+1=x^2−kx+1

or, kx = −kx

or, kx/x=−kx/x, (x≠0)

or, k=−k

or, k+k=0

or, 2k=0

or, k=0/2

Therefore: k=0

For all values of x, k = 0 is the only solution.
(Answer)