Is x^2 - 2x
1 to 1?
I keep getting yes every way I try to work it. Book says no. In the end, I get (with x1=1 and x2=-1 <- I use these numbers because of the x^2) f(x1) = -1 and f(x2)=3. I can't seem to find if x1 = x2
(btw, I know that by the graph it is not, but I am trying to work it out the books way)
I should add that my prof said in the notes that if the two answers to the two functions are the same, then it is not 1 to 1 BECAUSE they have the same answer. That's what has me confused.
One-to-one is usually associated with an interval. If the interval is not mentioned, we will assume that it is (-∞∞).
To find out if a function
f(x) is one-to-one on the interval, we need to know if it is possible to find different values of x1 and x2 for which f(x1)=f(x2), where x1-x2≠0.
If f(x1)=f(x2) then f(x) is not one-to-one on the given interval.
For example,
f(x)=x² on the interval (-∞,∞),
we can find
f(-1)=f(1), or f(-2)=f(2), therefore f(x)=x² is NOT one-to-one.
The horizontal line test says that if you can draw a horizontal line and intersect the function at two or more points, the function is NOT one-to-one. On the other hand, if it is impossible to do so, the function is one-to-one.
Try out your problem and post if you have other questions.
To determine whether the function is one-to-one, we need to check if different inputs yield different outputs. In other words, we need to find whether there are any two distinct values of x that produce the same value for f(x) = x^2 - 2x.
Let's solve the equation f(x) = f(y) and see if we can find any solutions for x and y where x ≠ y:
x^2 - 2x = y^2 - 2y
To simplify this equation, let's move all terms to the left side:
x^2 - y^2 - 2x + 2y = 0
Now, we can factor the left side of the equation:
(x - y)(x + y) - 2(x - y) = 0
Next, we can further simplify this equation by factoring out (x - y):
(x - y)(x + y - 2) = 0
To solve this equation, we have two cases:
Case 1: (x - y) = 0
If (x - y) = 0, it means that x = y. So, in this case, we have found a solution where x1 = x2.
Case 2: (x + y - 2) = 0
If (x + y - 2) = 0, this implies that x + y = 2. However, this equation does not specify a unique relationship between x and y.
Based on our analysis, we have found a solution (x1 = x2) for case 1, but there are no distinct x and y such that f(x) = f(y) for case 2.
Therefore, the function f(x) = x^2 - 2x is not one-to-one, as there exists a case where different inputs (x and y) yield the same output (f(x) = f(y)).