Posted by Laurie on Wednesday, June 24, 2009 at 8:58pm.
What are the choices?
A. f is an odd function
B. f is an even function
C. f is neither odd nor even
D.f is both odd and even
Totally confused and would appreciate some insight. Thanks!
an even function is one where f(x) = f(-x), that is, using either a positive number as input to the function you get the same result as if you used the opposite of that number
and odd function is where f(x) = -f(x)
so what if you tried f(5) and f(-5)
f(5) = 1 - 3(5)^2 = 1 - 75 = -74
f(-5) = 1 - 3(-5)^2 = 1 - 75 = -74
will that be true for all ± x's ?
mmmhhh?
another way to look at it:
an even function results in a reflection in the y-axis
an odd function results in a reflection in the origin.
So I would say that the answer would be that f can be both odd and even then. Am I correct?
NO, no, no
it is EVEN.
Just make a quick sketch of the graph.
Did you not read my explanations ?
Yes I read them, but misinterpreted them I suppose.
So this is an even function because x is being represented rather than y and this would be an even function resulting in a reflection in the y-axis?
I will sketch it out.
If this was a -f(x) then it would be odd? I think I understand now..Am I correct?
Related Questions
Algebra - Let f(x) = 1 – 3x^2. Which of the following is true? Please give ...
Algebra - Let f(x) = 1 – 3x^2. Which of the following is true? Give a brief...
Algebra - Let f(x) = 1 – 3x^2. Which of the following is true? Give a brief...
Physics-Rutherford - Can you please give me a brief explanation of Rutherfords ...
computer science - This week you will determine the time complexity for a game ...
Ashford - This week you will analyze the computational complexity of the game ...
Algebra - Determine the zeros of the function 3x^2 -8=10x. I know I rewrite it ...
chemistry - Can anyone give me a brief explanation about redox? I'm just not...
Algebra - Can someone please break down how to solve this equation. I am so ...
chemistry - (i) Give a brief explaination of the two bonding factors which ...
For Further Reading
