- #1

mmlm01

- 5

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, the horizontal line test for a continuous function must have a slope that is constant, there cannot be any maxima or minima, and the first derivative must always have the same sign.

- #1

mmlm01

- 5

- 0

Physics news on Phys.org

- #2

StatusX

Homework Helper

- 2,570

- 2

- #3

mmlm01

- 5

- 0

I think that the slope must be constant (positive or negative) in order for a function to pass the horizontal line test. I mean, it can be positive AND negative. I also believe that there cannot be any maxima or minima.

I don't want to seem unintelligent, but I am not so great at calculus to begin with, and my teacher this semester if fresh out of grad school. He is a really nice person, but he doesn't always seem to understand that we don't know everything he does. So, on some topics his instruction is sketchy at best. But, there's nothing I can do about that, so I am making the best of this situation.

- #4

StatusX

Homework Helper

- 2,570

- 2

- #5

mmlm01

- 5

- 0

StatusX said:So just make sure the first derivative of your function always has the same sign.

I am not sure what you mean by this. The 1st derivative of my function is (I think) f'(x)=sin x .

I know f(x)= x^2 does not have an inverse. The first derivative here is f'(x)=2x

I know f(x)=e^x does have an inverse. The first derivative here is f'(x)=lnx.

I just don't know how all of this is related. Is there a rule (or fundamental concept) I am missing?

- #6

StatusX

Homework Helper

- 2,570

- 2

Your derivative is wrong. Don't forget the derivative of x is 1. And the derivative of e^x is e^x. x^2 doesn't have an inverse because its derivative is negative for x<0 and positive for x>0. e^x does because it is always positive.

It's pretty intuitive if you think about what happens when a derivative changes sign. It means there is a maximum or minimum, and the function will fail the horizontal line test.

It's pretty intuitive if you think about what happens when a derivative changes sign. It means there is a maximum or minimum, and the function will fail the horizontal line test.

Last edited:

- #7

what

- 91

- 0

f'(x)= 1- sin(x) Can you figure out why this is either always positive or negative?

- #8

jdavel

- 617

- 1

"the derivative of your function is simple enough to compute

f'(x)= 1- sin(x) Can you figure out why this is either always positive or negative?"

I can't, can you?

- #9

what

- 91

- 0

yes if you wish to know the answer send me a message, I thought that it would be too easy for him if i just give him the answer.jdavel said:

"the derivative of your function is simple enough to compute

f'(x)= 1- sin(x) Can you figure out why this is either always positive or negative?"

I can't, can you?

- #10

jdavel

- 617

- 1

Weren't there some bounds given for the domain of f(x). If not, you're going to have a hard time showing the f(x) has an inverse!

- #11

mmlm01

- 5

- 0

Also, I am not 100% sure why f'(x)= 1- sin(x) is always positive or negative. Perhaps I have not had enough experience with derivatives for this to be evident immediately.

- #12

StatusX

Homework Helper

- 2,570

- 2

-sin x is always greater than or equal to -1. now add 1 to it.

- #13

what

- 91

- 0

- #14

jdavel

- 617

- 1

what,

"...0 to 2 is always positive."

What?

"...0 to 2 is always positive."

What?

- #15

StatusX

Homework Helper

- 2,570

- 2

- #16

jdavel

- 617

- 1

"The only thing that happens is that the inverse isn't differentiable at these points. But it's (x^1/3) a valid function."

True, but that's because the x^3 has an inflection point at x=0 rather than a local max or min. But the function in questioin, f(x) = x + cos(x), has lots of local maxs and mins. So what does that say about whether it has an inverse?

- #17

StatusX

Homework Helper

- 2,570

- 2

- #18

AKG

Science Advisor

Homework Helper

- 2,567

- 4

What does it mean to be 1-1 (one to one)? It means that two different x values, x1 and x2, will not give the same value f(x1) = f(x2). And think about the horizontal line test, isn't that exactly what it's about? So, without a graph, can you show that there are no two different x values for which the function takes the same value? Suppose f(a) = f(b), i.e. f takes on the same value at a and at b.

a + cos(a) = b + cos(b)

[cos(b) - cos(a)]/(b - a) = -1

This is to say that you can draw a secant through the curve (x, cos(x)) with slope -1, intersecting the curve at points (a, cos(a)) and (b, cos(b)). Suppose we have a pair of points through which the secant is less than -1. Then by mean value theorem, we'd have a place where cos(x) had a slope of less than -1, which is impossible, so -1 is the smallest possible value for a secant. Choose some c such that a < c < b, and we know that:

[cos(c) - cos(a)]/(c - a)

Suppose it is greater than -1, then [cos(b) - cos(c)]/(b - c) < -1, which would imply there is some d in (c, b) with cos'(d) < -1, which is impossible, so it cannot be greater than -1. Suppose it is -1, then [cos(b) - cos(c)]/(b - c) = -1. Then there is some d in (c, b) with cos'(d) = -1, and some e in (a, c) with cos'(e) = -1. but this requires d - e = 2kpi, for some positive integer k. But then b - a > 2kpi, so:

cos(b) - cos(a) < 2kpi, which is impossible because cos(b) - cos(a)

[cos(b) - cos(a)]/(b - a) = -1

hence our function is 1-1, and hence invertible.

- #19

jdavel

- 617

- 1

StatusX and what,

You are right! Adding the x term is just enough keep f(x) from ever turning back down. I even graphed it on Xcel, but to get more cycles, I used sin(10x). The x term couldn't keep that one from going down, so I was sure I was right.

Anyway, sorry about being so adamant. I'll try to be more careful.

jdl

Edit: This also applies to AKG. Didn't see your post until mine was up.

You are right! Adding the x term is just enough keep f(x) from ever turning back down. I even graphed it on Xcel, but to get more cycles, I used sin(10x). The x term couldn't keep that one from going down, so I was sure I was right.

Anyway, sorry about being so adamant. I'll try to be more careful.

jdl

Edit: This also applies to AKG. Didn't see your post until mine was up.

Last edited:

In order to prove that a function has an inverse, you must show that the function is both one-to-one (injective) and onto (surjective). This means that for every input, there is only one corresponding output, and that every output has at least one corresponding input. In other words, the function must pass both the horizontal line test and the vertical line test.

No, a function can only have one inverse. If a function has more than one inverse, then it is not a function. This is because a function must have a unique output for every input, and having multiple inverses would mean that there are multiple outputs for the same input.

To find the inverse of a function algebraically, you need to switch the x and y variables and solve for y. In other words, if the original function is f(x), the inverse function would be written as f^{-1}(x). Algebraically, this would look like y = f(x) and x = f^{-1}(y), and then solving for y in terms of x.

The process for showing that f(x) = x + cos x has an inverse involves first proving that the function is one-to-one and onto. This can be done by showing that the function is continuous and differentiable, and by using the horizontal line test and the vertical line test. Once it is established that the function has an inverse, you can find the inverse algebraically by switching the x and y variables and solving for y.

Yes, it is possible for a function to have an inverse without using a graph. As mentioned earlier, you can prove that a function has an inverse by showing that it is both one-to-one and onto. This can be done using various mathematical techniques, such as the horizontal and vertical line tests, without the use of a graph.

- Replies
- 2

- Views
- 5K

- Replies
- 4

- Views
- 1K

- Replies
- 10

- Views
- 4K

- Replies
- 3

- Views
- 1K

- Replies
- 12

- Views
- 1K

- Replies
- 6

- Views
- 1K

- Replies
- 2

- Views
- 2K

- Replies
- 1

- Views
- 1K

- Replies
- 6

- Views
- 1K

- Replies
- 4

- Views
- 3K

Share: