Asked by Lucy
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...
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...
Answers
Answered by
Damon
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
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
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