factor the numerator.
(2x^n -1)(2x^n+1)
dividing, leaving you with 2x^n+1
Well, 2(any number) is even, add one, it is always odd.
Simplify: ((2x^n)^2 - 1)/(2x^n-1)
where x is an integer and n is a positive interger.
for this one igot the answer the same as 2n^x+1, but will the value of the given expression be even, odd, or either? Please explain. :) Thank Yoo
6 answers
How do you get from x^n to n^x?
2x^n - 1
will be even if n is even, but will never be odd, because 1 = x^0 is an even power.
f(-x) = 2(-x)^n - 1 = 2x^n-1 = f(x) if n is even
f(-x) = 2(-x)^n - 1 = -2x^n-1 ≠ -f(x) if n is odd
2x^n - 1
will be even if n is even, but will never be odd, because 1 = x^0 is an even power.
f(-x) = 2(-x)^n - 1 = 2x^n-1 = f(x) if n is even
f(-x) = 2(-x)^n - 1 = -2x^n-1 ≠ -f(x) if n is odd
Oops. bobpursley got it right. I was reading it wrong, think about even or odd functions, not the value of the expression. :-(
Thank you guys so so so so much! I have been looking at this forever trying to see how to solve it, when it was right in front of me!! :D
But Steve, you used the equation 2x^n-1 rather than 2x^n+1, does that make a difference??
my bad - typo.
it makes no difference to the even/odd results.
it makes no difference to the even/odd results.
1 answer