#1
The table of values is consistent with an even function. They don't provide f(x) and f(-x) for any value of x.
Your answer is correct, because
f(-1) - f(2) = f(1)-f(-2) = -5-5 = -10
#2
The table is ok, for the same reason as given for #1.
f(0) + f(–0.53) = f(0)-f(0.53) = 2-2 = 0
Could someone tell me if I'm right, and if not what the correct answer is? Thank you! I'd appreciate an explanation, too, if you could. :)
1. Suppose you're given the following table of values for the function f(x),and you're told that the function is even:
x f(x)
-------------
–2 5
–0.35 –3
0 2
0.53 2
1 –5
Then:
f(2)= –5
f(.35) +f(–.53) = 1
***f(–1) – f(2) = –10
f(0) +f(–.53) = 0
Something is wrong. Given the table of values, the function can't be even.
2. Suppose you're given the following table of values for the function f(x), and you're told that the function is odd:
x f(x)
---------------
–2 5
–0.35 –3
0 2
0.53 2
1 –5
Then:
f(2)= 5
f(0.35) + f(–0.53) = –1
f(–1) – f(2) = –10
f(0) + f(–0.53) = 0
***Something is wrong. Given the table of values, the function can't be odd.
2 answers
f(-1)=5 and f(-2)=-5 so 5--5 = 5+5= 10 that is how it is even