If f(x) is an even function and g(x) is an odd function, which of the following must be even

Question

I. f(g(x))

II. f(x) + g(x)

III. f(x)g(x)

A. I only

B. II only

C. I and II only

D. II and III only

E. I, II, and III

Answer 1

Hi Woojin,

From the question, we know that for all n:

f(x) = even function, therefore... f(x) = f(-x) = x²ⁿ

g(x) = odd function, therefore... g(x) = -g(x) = x²ⁿ⁺¹

Looking at the options:

I. f[g(x)] then would be equal to f[-g(x)]

--> This is must be even because f(x) = f(-x), where in this case x would be "g(x)"

II. f(x) + g(x)...

x²ⁿ + x²ⁿ⁺¹ = 2x²ⁿ + x

Therefore x is always an odd function.

III. f(x)g(x)

In this case, we have one odd function multiplied by an even function, so...

Let f(x) = x²ⁿ and g(x) = x²ⁿ⁺¹ for all n.

Then f(x)g(x) = x²ⁿx²ⁿ⁺¹ = x^{(4n^2)+1}

--> f(x)g(x) is always an odd function.

Keep in mind you could do Case 1 with the 2n and 2n+1 exponents way, but it would be much more arduous.

Hope this helps!

-Winn