arildno said:Your attempt is wrong. f(x)=x is one-to-one, so is g(x)=1-x. But the sum f+g is not one-to-one.
Try to answer the following:
If a function is NOT one-to-one, what do you know occasionally occurs with the function that cannot happen with a one-to-one function?
oli4 said:Hi quadreg, have you studied derivatives yet ?
Cheers...
oli4 said:Exactly, yo've got it :)
To prove that f(x) is a one-to-one function, we need to show that for every input x, there is a unique output f(x). This can be done by using the horizontal line test, where we draw a horizontal line through the graph of f(x) and check if it intersects the graph at more than one point. If it does not, then f(x) is a one-to-one function.
Yes, the derivative can help in proving that f(x) is a one-to-one function. If the derivative of f(x) is always positive or always negative, then f(x) is a strictly increasing or strictly decreasing function, which implies that it is one-to-one.
The domain of f(x) is all real numbers, as there are no restrictions on the values of x that can be plugged into the function.
Yes, f(x) is a continuous function as it is a polynomial and all polynomials are continuous.
Yes, algebraic manipulation can also be used to prove that f(x) is a one-to-one function. We can start with the assumption that f(x) is not one-to-one and then use algebra to show that this leads to a contradiction, proving that f(x) must be one-to-one.