Posted by **xavier** on Wednesday, October 3, 2012 at 11:43pm.

Auppose f is even but g is an arbitrary function (possibly neither even nor odd).

(a) Is f o g still even? Show why or why not.

(b) Is g o f even? Show why or why not.

- Math -
**Steve**, Thursday, October 4, 2012 at 10:35am
fog is even, since f is even.

since f(-x) = f(x), f(-g) = f(g)

gof is unknown. for example,

f(x) = x^2

g(x) = x

g(f) = f = x^2 is even

g(x) = √x

g(f) = √f = x is odd

g(x) = √x + 1

g(f) = √f + 1 = x+1 neither even nor odd

## Answer this Question

## Related Questions

- math - I know It's probably an easy question but I don't know remember how to do...
- odd and even functions - Using f is odd if f(-x) = -f(x) or even if f(-x) = f(x...
- Algebra - Let f(x) = 1 – 3x^2. Which of the following is true? Give a brief ...
- Algebra - Confused Please Help! Thanks! Let f(x) = 1 – 3x^2. Which of the ...
- Algebra - Let f(x) = 1 – 3x^2. Which of the following is true? Please give us a ...
- Algebra - Let f(x) = 1 – 3x^2. Which of the following is true? Give a brief ...
- pre cal - Determine algebraically whether the function is even, odd, or neither ...
- discrete math - prove that if n is an integer and 3n+2 is even, then n is even ...
- Math (precalculus) - I have a question about the symmetry of graphs, but maybe ...
- Calculus, check my answer, please! 1 - Did I get this practice question right? 1...

More Related Questions