Name: Anonymous 2009-09-13 17:13
So I'm going back to the basics and the book gives an example stating that if I want to show f(x)= 3x-5 is injective I do the following:
f(x1)= f(x2)
3x1-5=3x2-5
3x1=3x2
x1=x2....thus it is injective.
Now let's say we didn't know the following is a parabola and thus could not apply the horizontal line test. f(x)=x^2
x^2(1)=x^2(2), then we take the square root of both sides so we have..
x1=x2.
According to that the parabola is one-to-one, but we know it isn't. What gives?
f(x1)= f(x2)
3x1-5=3x2-5
3x1=3x2
x1=x2....thus it is injective.
Now let's say we didn't know the following is a parabola and thus could not apply the horizontal line test. f(x)=x^2
x^2(1)=x^2(2), then we take the square root of both sides so we have..
x1=x2.
According to that the parabola is one-to-one, but we know it isn't. What gives?