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August 1, 2014

August 1, 2014

Posted by **Samantha** on Wednesday, October 2, 2013 at 11:11pm.

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.

- Calculus, check my answers, please! -
**Steve**, Wednesday, October 2, 2013 at 11:58pm#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

- Calculus, check my answers, please! -
**holly**, Thursday, October 3, 2013 at 3:43pmf(-1)=5 and f(-2)=-5 so 5--5 = 5+5= 10 that is how it is even

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