Answer:
a. even
b. odd
c. odd
d. even
Step-by-step explanation:
We suppose that "prove algebraically" means that you are to substitute -x for x in each equation and show that f(x) = -f(-x) for an odd function, and that f(x) = f(-x) for an even function.
It is easier to do this generically, so that you can make use of the rule that ...
- if a polynomial function consists only of even-degree terms, it is an even function
- if a polynomial function consists only of odd-degree terms, it is an odd function
__
So, for even-degree, we can assume that the polynomial is of the form ...
f(x) = ax^2 + b
Then we have ...
f(-x) = a(-x)^2 +b = (-1)^2·ax^2 +b = ax^2 +b
or ...
f(-x) = f(x) . . . . regardless of the values of "a" and "b"
__
For an odd-degree function, we can assume that the polynomial is of the form ...
g(x) = x·f(x) . . . . where f(x) is of even degree
Then ...
g(-x) = -x·f(-x) = -x·f(x) . . . . since f(x) = f(-x) as shown above
or
g(-x) = -g(x) . . . . . . . . regardless of the coefficients in f(x)
__
a. f(x) matches the form of our even-degree f(x) above, so is even.
b. g(x) matches the form of our odd-degree g(x) above (with a=0), so is odd.
c. h(x) = x(5x^2 -1) matches the form of our odd-degree g(x), so is odd.
d. f(x) matches the form of our even-degree f(x) above, so is even.
_____
Comment on the problem
For part (b), we have assumed that "x +1x" means 2x. If it is supposed to be something else, it needs to be properly written. (x+1/x is an odd function, because -x+1/(-x) = -(x +1/x).)
