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If g(x) = 3 + x + e^x, find g^-1(4)

by illjazz
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illjazz
#1
Jul28-08, 04:16 AM
P: 59
1. The problem statement, all variables and given/known data
If [tex]g(x) = 3 + x + e^x[/tex], find [tex]g^{-1}(4)[/tex]


2. Relevant equations
Not sure. The log laws don't seem to apply. Probably laws/rules related to the number e.


3. The attempt at a solution
So I know the whole process and technically have this solved, but not because I understand how to do it but because I have it written down from a short discussion in class. It goes like this:

[tex]g(x)=3+x+e^x[/tex]

[tex]y=3+x+e^x[/tex]

[tex]y-3=x+e^x[/tex]

And this is where I get stuck. How do I "factor" (x+e^x)? Alas, what I have continues:

[tex]2x=ln(y-3)[/tex]

[tex]x=\frac{ln(y-3)}{2}[/tex]

Then,

[tex]g^{-1}(x)=y=\frac{ln(x-3)}{2}[/tex]

and

[tex]g^{-1}(4)=y=\frac{ln(4-3)}{2}[/tex]

[tex]=\frac{ln(1)}{2}[/tex]

[tex]=\frac{0}{2}[/tex]

[tex]=0[/tex]

What I don't understand is how to get from (e^x + x) to 2x.

TIA
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arildno
#2
Jul28-08, 04:28 AM
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First off, let us indtroduce y as :
[tex]y=3+x+e^{x}[/tex]
Now, g^{-1} is the INVERSE function of g, so that [itex]g^{-1}(4)[/itex] is that value x has whenever y has the value 4

But, this is the same as finding a solution to the following equation:
[tex]4=3+x+e^{x}[/tex]
Do you agree?

Furthermore, make a valid argument that the solution x cannot be strictly greater than zero.
illjazz
#3
Jul28-08, 04:37 AM
P: 59
Quote Quote by arildno View Post
First off, let us indtroduce y as :
[tex]y=3+x+e^{x}[/tex]
Now, g^{-1} is the INVERSE function of g, so that [itex]g^{-1}(4)[/itex] is that value x has whenever y has the value 4

But, this is the same as finding a solution to the following equation:
[tex]4=3+x+e^{x}[/tex]
Do you agree?

Furthermore, make a valid argument that the solution x cannot be strictly greater than zero.
I agree completely. You'll see that I solved this one all the way above, however.. I only had one particular problem at one specific step of finding the inverse function. I suppose I could have saved a bit of writing by solving for 4 = 3+x+e^x but what I did above is how my book said to do it..

1. Replace "f(x)" with y
2. Solve for x
3. Switch y and x

And then with the result, I plugged in 4 into the inverse. In other words, what you suggest and what I did are equivalents.

But yeah.. all I'd love to know is how they get from

[tex]e^x+x[/tex] to 2x.

Thanks!

arildno
#4
Jul28-08, 04:57 AM
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If g(x) = 3 + x + e^x, find g^-1(4)

They don't, since it is wrong. You won't be able to find an explicit functional form of the g^{-1} function, but you can nonetheless determine that when y is 4, then x is 0.
HallsofIvy
#5
Jul28-08, 05:05 AM
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You've misunderstood the problem. As arildno says, you can't find a general formula for the inverse function but it is easy to solve 3+ x+ ex= 4 "by inspection".
illjazz
#6
Jul28-08, 05:12 AM
P: 59
Quote Quote by arildno View Post
They don't, since it is wrong. You won't be able to find an explicit functional form of the g^{-1} function, but you can nonetheless determine that when y is 4, then x is 0.
Phew.. that's a relief. I was wrecking my brain over what I'd missed there.. ok, so why would my Calc prof do that step? Specifically, I mean the two lines right before I said "And this is where I get stuck..." in my original post. I find it hard to believe that that's completely arbitrary.. the 2x that is. There must be something to it, no?

Also, I may have spoken too fast when I said I agreed. I of course know you're right, but here's what trips me up: 3+x+e^x is the ORIGINAL function. So when I solve for 4 = 3+x+e^x, I'm solving for y = 4 using the original function and NOT the inverse. What the question asks, though, is to find the value of [tex]g^{-1}(4)[/tex], so how is it possible that we can let 4 equal the original function and yet still get the value of the inverse function at 4?

I'll give this a shot:

[tex]4=3+x+e^x[/tex]

[tex]1=x+e^x[/tex]

And now what?
arildno
#7
Jul28-08, 05:41 AM
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Quote Quote by illjazz View Post
Phew.. that's a relief. I was wrecking my brain over what I'd missed there.. ok, so why would my Calc prof do that step? Specifically, I mean the two lines right before I said "And this is where I get stuck..." in my original post. I find it hard to believe that that's completely arbitrary.. the 2x that is. There must be something to it, no?
If your calculus professor said this, which I strongly doubt, he shouldn't be a calculus professor.
Are you sure you haven't missed out something major in your notes?
Also, I may have spoken too fast when I said I agreed. I of course know you're right, but here's what trips me up: 3+x+e^x is the ORIGINAL function. So when I solve for 4 = 3+x+e^x, I'm solving for y = 4 using the original function and NOT the inverse.

Let's keep this simple.

Let y=f(x)=2x.
Now, obviously, x=y/2, or, equivalently,[itex]x=f^{-1}(y)=y/2[/itex]

Let us now compute [itex]f^{-1}(3)[/itex] from what you call the original equation:
3=2x, that is x=3/2.

Let us also compute f^{-1}(3) from its explicit functional form y/2: Then we get, lo and behold! f^{-1}(3)=3/2

Do you understand that?
illjazz
#8
Jul28-08, 05:47 AM
P: 59
Quote Quote by arildno View Post
If your calculus professor said this, which I strongly doubt, he shouldn't be a calculus professor.
Are you sure you haven't missed out something major in your notes?
Well I wouldn't make that up, would I? What would I have to gain from it? That whole path of solving the problem I took straight out of my Calc notebook, and everything in there comes straight off the blackboard in class. He discussed the problem because there was a group of people in the course who wanted him to go over that specific problem. Perhaps he tried to simplify something by using that "2x" term? I honestly don't know. What I do know is that that's what he wrote on the board.

Quote Quote by arildno View Post
Let's keep this simple.

Let y=f(x)=2x.
Now, obviously, x=y/2, or, equivalently,[itex]x=f^{-1}(y)=y/2[/itex]

Let us now compute [itex]f^{-1}(3)[/itex] from what you call the original equation:
3=2x, that is x=3/2.

Let us also compute f^{-1}(3) from its explicit functional form y/2: Then we get, lo and behold! f^{-1}(3)=3/2

Do you understand that?
Gotcha! Yeah that makes sense. Thanks!

I still don't see what my next step would be from the last line of my previous post though :/
arildno
#9
Jul28-08, 06:00 AM
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Quote Quote by illjazz View Post
Well I wouldn't make that up, would I? What would I have to gain from it? That whole path of solving the problem I took straight out of my Calc notebook, and everything in there comes straight off the blackboard in class. He discussed the problem because there was a group of people in the course who wanted him to go over that specific problem. Perhaps he tried to simplify something by using that "2x" term? I honestly don't know. What I do know is that that's what he wrote on the board.
I'm not accusing you of "making it up".
But, you should know that myself, and many others on this board have, or have had lots of professional teaching experience.
And I can tell you that the number one mistake among all students (not just you!) are not to take correct notes, or jumble them together into something unrecognizable.
The students haven't made it up, they have made a mistake, which is something very different.
This is extremely common, and we see this also very frequently on these boards.

I advise you to speak with a few of your fellow students, and then your lecturer, in order to find the source of this confusion.

It DOES happen that a professor writes complete nonsense on the blackboard; usually, that is confined to the philosophy department, but at times, even the maths department gets infected, too...
Gotcha! Yeah that makes sense. Thanks!

I still don't see what my next step would be from the last line of my previous post though :/
Okay, we have:
[tex]1=x+e^{x}[/tex]
Suppose that x>0 (x is a positive number).
Will e^x then be greater, equal or less than 1?
And how does that affect the possibility that the solution x might be a strictly positive number?
spideyunlimit
#10
Jul28-08, 06:21 AM
P: 61
well e^x = 1 + x + x^2/2! + x^3/3! ..... + x^n/n!
so in e^x + x , they added one x from the euler series and one from the expression, to get 2x... dunno wht they did with the rest and how they got the next step!
arildno
#11
Jul28-08, 06:52 AM
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Quote Quote by spideyunlimit View Post
well e^x = 1 + x + x^2/2! + x^3/3! ..... + x^n/n!
so in e^x + x , they added one x from the euler series and one from the expression, to get 2x... dunno wht they did with the rest and how they got the next step!
Which merely shows that it is incorrect to substitute the expression with 2x, as tstated above.
spideyunlimit
#12
Jul28-08, 07:01 AM
P: 61
Yea, that's what I'm trying to figure out, what they did.
spideyunlimit
#13
Jul28-08, 07:02 AM
P: 61
Btw, for inseparable functions, there is a different method too, where you don't have to completely represent the equation in terms of x and swap... I'll post the method when i remember it. :)
illjazz
#14
Jul28-08, 07:25 AM
P: 59
Quote Quote by arildno View Post
I'm not accusing you of "making it up".
But, you should know that myself, and many others on this board have, or have had lots of professional teaching experience.
And I can tell you that the number one mistake among all students (not just you!) are not to take correct notes, or jumble them together into something unrecognizable.
The students haven't made it up, they have made a mistake, which is something very different.
This is extremely common, and we see this also very frequently on these boards.

I advise you to speak with a few of your fellow students, and then your lecturer, in order to find the source of this confusion.

It DOES happen that a professor writes complete nonsense on the blackboard; usually, that is confined to the philosophy department, but at times, even the maths department gets infected, too...


Okay, we have:
[tex]1=x+e^{x}[/tex]
Suppose that x>0 (x is a positive number).
Will e^x then be greater, equal or less than 1?
And how does that affect the possibility that the solution x might be a strictly positive number?
e^x will always be greater than 1 because its graph passes through (0,1) and keeps increasing from there. With the requirement x > 0 the solution to the equation when solved for x can only be positive for obvious reasons. The fact that e^x will always be positive (even when x < 0, in fact!) simply underlines the fact that when this function is solved for x, it can only be positive.

Still, how does any of this help me get the value of the inverse of the function when x = 4?

Also, I did not mean to imply that you accused me of making anything up. My apologies if it came across that way. I can only nod in agreement as a response to that paragraph--I understand. I don't know the last time I made a mistake copying anything off a board, so I'm pretty confident that I got it down right.. which made it all the more puzzling to me where that 2x came from.

I appreciate the help :)
arildno
#15
Jul28-08, 07:34 AM
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Quote Quote by illjazz View Post
e^x will always be greater than 1 because its graph passes through (0,1) and keeps increasing from there. With the requirement x > 0 the solution to the equation when solved for x can only be positive for obvious reasons. The fact that e^x will always be positive (even when x < 0, in fact!) simply underlines the fact that when this function is solved for x, it can only be positive.

Still, how does any of this help me get the value of the inverse of the function when x = 4?
Re-read closely what you FIRST wrote, before you lost yourself again:
e^x will always be greater than 1 because its graph passes through (0,1) and keeps increasing from there
What conclusions can be drawn from this?

I'll help you out:

Since, for positive x, e^x is greater than 1, therefore, the sum of x and e^x must always be greater than 1.
Thus, we can draw the conclusion:
There exists NO positive x which is the solution to the equation:
[tex]1=x+e^{x}[/tex]
since the expression on the right-hand-side is strictly GREATER than 1!

Do you understand that?

Now, try and make a similar argument where you prove that there cannot be any negative x-solution to that equation either!
illjazz
#16
Jul28-08, 07:58 AM
P: 59
Absolutely! I guess what threw me of is what I'd initially written.. ending up at equation = 0. Something not existing, aka being undefined, does not mean it is equal to 0, as we know..

I'll have to get back to you later on that "proof" because I'm working on a bunch of other problems as well.

What, then, is the value of [tex]g^{-1}(4)[/tex]?
arildno
#17
Jul28-08, 08:06 AM
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I've already told you earlier.
What I proposed to you was how you can gradually circle yourself down to the answer x=0 as the only solution by first eliminating the possibilities of strictly postive or strictly negative solutions. That leaves you with only one possibility, namely that x, if the equation has a solution must be 0, and by inspection, you'll confirm it is, indeed, a solution to the problem.
HallsofIvy
#18
Jul28-08, 09:02 AM
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What is the value of e0?


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