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StatusX

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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.

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StatusX

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I am not sure what you mean by this. The 1st derivative of my function is (I think) f'(x)=sin x .StatusX said:So just make sure the first derivative of your function always has the same sign.

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?

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StatusX

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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.

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

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"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?

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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?

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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!

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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.

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StatusX

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

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what,

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

What?

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

What?

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StatusX

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"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?

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StatusX

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AKG

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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.

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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.

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