I have a question about the symmetry of graphs, but maybe it's more of a simple factoring question...
Why is f(x)=x+(1/x) odd, while h(x)=x-x^2 is neither even nor odd?
I understand that f(-x)=-x-1/x=-(x+1/x)=-f(x) is odd because f(x)=f(-x).
Then for h(-x)=-x-x^2...why can't you factor out the negative in this question like the previous one? But my textbook leaves it like this and says it's neither even nor odd...
h(x) = x - x^2
h(-x) = -x -x^2
but x is not -x so that part is odd
while -x^2 is -x^2 so that part is even
for odd I needed
h(-x) = -x + x^2 = -(x-x^2)
for even I needed
h(-x) = x - x^2
To determine whether a function is even, odd, or neither, we need to examine how the function behaves under the process of reflection (inversion) about the y-axis and the origin.
Let's first analyze the function f(x) = x + (1/x). To check if it is odd, we substitute -x in place of x and see if the resulting expression is equal to the negation of the original function.
f(-x) = (-x) + (1/(-x)) = -x - (1/x)
Now, let's compare f(-x) with -f(x):
-f(x) = -[x + (1/x)] = -x - (1/x)
You can see that f(-x) is equal to -f(x). Hence, f(x) = x + (1/x) is an odd function.
Now, let's analyze the function h(x) = x - x^2. To determine whether it is even, we substitute -x in place of x and see if the resulting expression is equal to the original function.
h(-x) = (-x) - (-x)^2 = -x - x^2
By comparing h(-x) with h(x), we can observe that h(-x) is not equal to h(x). Therefore, h(x) = x - x^2 is not an even function.
To check if h(x) is odd, we substitute -x in place of x and see if the resulting expression is equal to the negation of the original function.
-h(x) = -[x - x^2] = -x + x^2
Again, by comparing -h(x) with h(-x), we find that they are not equivalent. Therefore, h(x) is not odd either.
In summary, f(x) = x + (1/x) is odd since f(-x) = -f(x), while h(x) = x - x^2 is neither even nor odd since h(-x) does not equal h(x) and -h(x) does not equal h(-x).