Curve sketching y=e^f(x) (1 Viewer)

[martin]

Looking at the 2003 question (attached) I don't think that e^f(x) would have an inflexion point.

As x->1 from the right, f(x)->-inf so e^f(x)->0. Then as we go further to the right x->inf so f(x)->0 from below, so e^f(x)->1 from below. I think you just start (with an open circle) at (1,0) and go smoothly to the asymptote y=1 concave down.

To back this up look at a plot of f(x)=-2/(x^2-1). This looks a lot like the picture in the exam. Then graph e^f(x) and it doesn't have an inflexion.

Or find the second derivative when x>1 and show it's negative so the graph is concave down.

Edit: After reading a few posts I agree that it could have a point of inflexion, but it doesn't look like it would to me.

 

[who_loves_maths]

martin, thanks for posting that graph up

you can deduce from that graph that e^(f(x)) for that particular y=f(x) cannot have a point of inflexion. from general integration, any POI can only occur in region where y is concave down so for 'x' between -1 and 1 this region of graph is immediately irrelevant. As x---> infinity, the graph approaches a horizontal asymptote, so dy/dx ---> 0 (but never gets there), so dy/dx = 0 is in turn a horizontal asymptote of the graph of dy/dx as x---> infinity. now, the second derivative (rather than thinking of it as a tester for POI) is by definition the gradient function of the first derivative; so as dy/dx approaches a horizontal asymptote for x--->infinity, then the gradient of dy/dx near infinity approaches (but never gets to) 0, ie. d^2y/dx^2 ---> 0 for x---> infinity. now for POI to occur on e^(f(x)), then 0 = f''(x) + (f'(x))^2, but both f''(x) and f'(x) tend toward zero at the same rate as 'x' gets larger so, (f'(x))^2 will be significantly smaller (absolute value) than the absolute value of f'(x) or f''(x), so f''(x) + (f'(x))^2 will not be zero. so there is no inflection point on e^(f(x)) for that particular graph of f(x).
the same is applicable to the other side: x---> -infinity, since the graph is symmetric.

also it won't make sense to have a POI occurring as x--->+/- 1, since if a POI exists near x = +/- 1 then it implies a concavity change and so the graph is (locally) smooth or 'even' (ie. not steep or asymptotic) for greater values extending beyond either side of x=+/- 1. but this cannot be the case since the graph is in fact discontinuous at x=+/- 1, so the graph of e^(f(x)) as x--->+/- 1 would be asymptotic and steep near +/- 1, which then defies the (logical) implication of the existence of any POI near there.

 

[martin]

I agree that you only have to look at x>1 but I think that it is possible to have a point of inflexion there. I don't really follow your argument who_loves_maths so I've made an example.

Consider the function that is equal to 2ln(x-1) for x between 1 and 1.2, equals -2/(x^2-1) for x bigger than 1.4 and is smooth in between. Attached is graph f(x) which looks like the exam question.

Then e^f(x) is attached and has to have a point of inflexion.

The lesson to be learnt is that as was shown earlier in this thread, a graph of f(x) can't tell you the concavity of e^f(x).

 

[no_arg]

Exactly!
Simple concave down increasing f can have multiple inflection points in e^f!!
Makes you appreciate the algebra!

 

[who_loves_maths]

I agree that you only have to look at x>1 but I think that it is possible to have a point of inflexion there. I don't really follow your argument who_loves_maths so I've made an example.

Consider the function that is equal to 2ln(x-1) for x between 1 and 1.2, equals -2/(x^2-1) for x bigger than 1.4 and is smooth in between. Attached is graph f(x) which looks like the exam question.

Then e^f(x) is attached and has to have a point of inflexion.

The lesson to be learnt is that as was shown earlier in this thread, a graph of f(x) can't tell you the concavity of e^f(x).

i see what you have done, but there is a fundamental difference between your graph and the graph given in the actual question. your graph is a piecemeal, in other words, you don't have a fixed function f(x) over the entire real 'x', also, you have not given a function that links the two parts of the graph for 'x' between 1.2 and 1.4, and as a result you can't claim as you did that

"and is smooth in between" for the simple reason that you don't know if it is smooth in between or not.
in fact, how can you be sure that there exists a function that describes a smooth curve connecting x=1.2 and 1.4 and at the same time will not make the points at x=1.2 and x=1.4 CRITICAL POINTS ?
(in fact, they probably are critical points in which case e^(f(x)) will have critical point at x=1.2 and 1.4 also meaning that e^(f(x)) is a piecemeal too)

i can see where you are coming from, but ultimately your result is purely cosmetic and very subtle. but if you think about it, you'll see that in your graph of e^(f(x)) there is no POI explicitly shown, ie. the POI must lie in the region (assuming the curve is smooth) between x=1.2 and x=1.4 ---> which is exactly the region that you do not have a function for. so you cannot be sure that a POI exist UNTIL you are sure that a function for that region exists and know what it is.
in other words, the flaw in your argument is your ASSUMPTION that there is indeed a smooth curve between 1.2 and 1.4 connecting the other two parts AND that x=1.2 and x=1.4 are not critical points in your piecemeal function; an assumption that you cannot prove. and since your mysterious POI must lie in this unknown region then you cannot even be sure of its existence.
[a simpler way of putting what i'm saying is that since your graph is a piecemeal function, then the graph of e^(f(x)) will also be piecemeal which means the region between x=1.2 and x=1.4 can be of a function that does not locally exhibit a POI in that region which means that overall the graph does not have a POI. (ie. you cannot be sure that the two parts of your graph are perhaps simply connected by a curve that doesn't have a POI on it in the "unknown" region.) ]

so since your original f(x) graph is a piecemeal, then so will your e^(f(x)) be one, which means it's not necessary for a POI to even exist since the graph is NOT smooth. (POI in your graph can only exist if you can prove no critical points exist and that e^(f(x)) is in fact smooth even though f(x) is a piecemeal ---> this is something i personally believe cannot happen, so i retain my argument in that no POI exists.)

i hope i made myself clear in what i just wrote. sorry if i'm bit repetitive martin but i hope you understand what i've tried to say.

P.S. even if your piecemeal is in fact smooth (which it's not) then it wouldn't be a piecemeal anyways, because you could generate a function in its totality to describe a completely smooth curve, and in this case your examples of using piecemeals to demonstrate your point will automatically break down. Also, i believe that if you can connect the two parts of the graph up smoothly then the global picture of the graph will not be monically increasing (which is what you are hoping because you are comparing this to your other HSC graph) which means it will be different to the HSC graph for x>1 and so that would not make a good comparison to the HSC graph, so i still retain my argument of no POI existing for that graph in my last post.

 

[Slidey]

Fee free to explore the use of calculus in solving this problem, but please don't pretend that the solution would be applicable to the thread starter's requirements, namely a solution which is viable under HSC examination conditions.

Yeh I am aware that you can find the inflection point when the second derivative is zero, but my problem is that you are not given the equation of f(x) in the first place to differentiate.

 

[martin]

As buchanan correctly points out I was wrong before. If we take f(x)=-2/(x^2-1) then e^f(x) has an inflexion point (I wasn't graphing with enough detail). See attachments.

So this example shows that it is possible to have an inflexion point. A modified version of my above example with 1/2*log(x-1) would show we could have a similar graph without an inflexion point.

As to the validity of the function I put up before: this is a subtle issue and reflects a misunderstanding that I think everybody has in high school. A function is not always a formula (technically it is just a set of ordered pairs for x and y coordinates). If I give you a graph of a function which has measurements on the axes (and you can measure infinitely finely) then you can find the y coordinate corresponding to any x coord. If the graph is continuous and differentiable (so it doesn't have any corners or jumps in it) then you can find the derivative just by taking the limit of rise/run as run->0.

So in my function above I actually specified an infinite number of different functions with the understanding that you could draw a nice one on the graph. If you wanted to fit a particular line to it you could, perhaps using a cubic spline which allows you to set the derivatives at the end points.

As to the relevance of this to the original question, of course you couldn't use calculus in the exam. But if we can show 2 particular examples of f(x) that look like the curve given (in the sense that they have the same turning points, inflexions, asymptotes) one of which has e^f(x) with inflexion and one has e^f(x) without inflexion then it shows that either is right.

 

[martin]

I say that a function that is equal to 1/2*ln(x-1) for x near 1 and then goes to the asymptote which is concave down, continuous and differentiable fits the bill. Attached is a drawing of one such function (if you can accept my poor freehand skills).

I'm not sure if there is a function with the desired property representable in terms of elementary functions like sin,log,polynomials.

 

[who_loves_maths]

hi martin and buchanan,

okay well after actually carefully reading most of the posts in this thread again i've discovered that i probably posted too hot-headedly on previous occasions without giving too much thought.
lol, i didn't actually think about the graph of the equation y=2/(1-x^2) until only now, and yes you're right it has POIs, in fact it has 2. but on the other hand, it is still only speculatory that that eqaution is in fact the equation for that HSC graph.

and about what you said buchanan:

BUT if you assume that f(x) is a rational function, then there must be an inflection for y=e^f(x). Note I'm not saying that this is true in general...

well this statement is certainly not true in general... BUT, let me just add this (and plz correct me once again if i am wrong):

assuming F(x) is a rational function such that F(x) = H(x)/G(x) , where H(x) and G(x) are polynomials of degree 'h' and 'g' respectively with 'h' as an integer larger than or equal to 0, and 'g' an integer larger than or equal to 1.
then:
1) If h>=g, the graph of e^(F(x)) will EITHER have NO POIs, OR, have POI that occur only in pairs where the NUMBER of pairs of POIs cannot exceed the value (g+h-1).

2) If h< g, AND, If (h+g) yields an ODD integer, then the graph of e^(F(x)) will have AT LEAST 1 POI. otherwise, the graph will have either no POI or have them occur in pairs.


applying this to the "original" HSC graph, which for the sake of application here I will assume to be the graph of y=2/(1-x^2), then h=0, and g= 2. ie. h< g, but (h+g) = 0 +2 = 2 = even number; therefore, the graph of e^(2/(1-x^2)) will have either NO POI or have them in pairs where the number of pairs cannot exceed (g+h-1) = 2 +0 -1 = 1. so immediately, it can be inferred that the graph has either no POI, OR, a pair of POIs.
(and since you guys know already that one POI exists, then there must be a second one too, so the graph in fact has a pair of POIs)

this also goes to answering your last "question" (if i can use that word) buchanan:

The challenge is now to present an example model f(x) which also looks like the curve drawn on the question paper for which ef(x) has no inflections (or doesn't go flat as x→1+)

to do so, it should be possible according to what i found above to find other rational functions (you don't need to look for fancy curves, which is a relief) that retains the same general shape and look of that HSC curve but for which there is no POI on e^(F(x)).

 

[no_arg]

What about the rational function f(x)=x/1? or (x-1)/(x-1)?

 

[no_arg]

It's interesting that it could also be a straight line!
That would be a brave move in an exam!

 

[underthebridge]

hmmm interesting example, and in that case Im sure you wouldnt get marked down for not including inflection points as i think buchanan pointed out at the top of the page. The answer that i saw with the inflection points was from those answer booklets the board of studies sends out for each past paper.