Question

Can someone explain what's one to one function is in relate to this b(x)=5x^3-7x?

Answer 1

A function is one to one if it passes the horizontal line test. In other words, it if you picture a horizontal line over your graph, it cannot intersect with the graph more than once. If it does, it fails the horizontal line test. This means it's inverse does not exist.

If you graph this function, you see that it fails the horizontal line test. Thus, it is not one-to-one.

Answer 2

That any vertical line will intersect a graph of the functions at only one point.

Answer 3

A function \( f:A \to B \) is said to be injection if different elements of \( A \) have different images in \(B \)

Answer 4

Note, to make my answer more rigorous, I'm assuming you already know your function is a function, meaning you know it passes the vertical line test.

Answer 5

yes, I do know that definition of a one to one function, but how would you answer it regards to the function above? Let's say what if I don't have a graphing calculator?

Answer 6

Horizontal line test is not necessary, say \( a \) and \( b \) be two arbitrary elements, then \( f(a)=f(b) \) doesn't provide unique solution \( a=b \) which indicates that this isn't injection.

Answer 7

Can we use calculus to show the function is one-to-one?

Answer 8

With most problems, to prove injectivity, you want to prove:\[f(x_1)=f(x_2)\Longrightarrow x_1 = x_2\]However, because of this particular function, that might not be so easy. Since the function is odd though, using calculus gives a really fast way to show the function is injective.

Answer 9

@Joe: \( 5a^3-7a=5b^3-7b \Rightarrow a(5a^2-7)=b(5b^2-7) \Rightarrow a=b \text{ and } a = \pm b\)

No unique solution hence not injection. Am I missing something?

and on using calculus, do you mean by plotting the function and then test for horizontal line test?

Answer 10

what you have is good. By using calculus, you get the first derivative, and show that its going to take on positive and negative values, so the actual graph (not the derivative) will go up and down, so its not injective.

Answer 11

Do you mean showing inflection points ?

Answer 12

Im just using the first derivative, so its critical points.

Answer 13

Hm cool.