1. ## help with computation

Compute [(2h + hg)/(g^2+1)] (0), where g is the function below
f(x) = x sin(1/x), x not = 0
f(x) = 0, x = 0

g(x) = xf(x).

Define h : R - R by
h(x) = {x^2, x rational
h(x) = {0, x irrational

I do not even understand the question, can anyone help

2. Originally Posted by 450081592
Compute [(2h + hg)/(g^2+1)] (0), where g is the function below
f(x) = x sin(1/x), x not = 0
f(x) = 0, x = 0

g(x) = xf(x).

I do not even understand the question, can anyone help
You haven't specified what $\displaystyle h(x)$ is.

Are you sure you aren't trying to find

$\displaystyle h(0) = \frac{2f(0) + g(0)f(0)}{g^2(0) + 1}$?

3. Originally Posted by Prove It
You haven't specified what $\displaystyle h(x)$ is.

Are you sure you aren't trying to find

$\displaystyle h(0) = \frac{2f(0) + g(0)f(0)}{g^2(0) + 1}$?
sorry, I forget to rewrite the function

Define h : R - R by
h(x) = {x^2, x rational
h(x) = {0, x irrational

I think we should use the quotient rule here, but I dont know how to start it