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

[underthebridge]

I have a question in regards to curves of the form y=e^f(x) where you are just shown a graph of y=f(x) (eg. 2003 HSC). I have noticed that occasionally there are inflexion points on these curves and I am not sure how to find out whether the curve will have inflexion points or not. I have worked out that if there are inflexion points on y=e^f(x), they will only occur where y=f(x) is concave down (2nd derivative is negative). I did this by differentiating y=e^f(x) twice, but it is not always the case that if f(x) is concave down y=e^f(x) will have inflexion points.

 

[Jumbo Cactuar]

i have that the conditions on are;

f'(x)² = - f''(x)
AND
f'''(x) + f'(x)³ not equal to 0

 

[who_loves_maths]

well clearly the only other possibility is that the inflection point occurs at the x value for which y=f(x) has a horizontal point of inflexion (eg. created by a triple zero) ie. first derivative = 0 = second derivative.
all other cases involve the second derivative being either positive or negative (0 is neither) and you have already taken that into account.

 

[who_loves_maths]

just forgot to mention in my last post:

like i said only other possiblity is if there's a horizontal P.O.I on y=f(x), WITH THE ADDITIONAL CONDITION THAT: third derivative is NOT equal to zero for the particular 'x' in question.

 

[Jumbo Cactuar]

We are talking y=e^f(x),..

all that has been resolved in my post.

 

[withoutaface]

I have a question in regards to curves of the form y=e^f(x) where you are just shown a graph of y=f(x) (eg. 2003 HSC). I have noticed that occasionally there are inflexion points on these curves and I am not sure how to find out whether the curve will have inflexion points or not. I have worked out that if there are inflexion points on y=e^f(x), they will only occur where y=f(x) is concave down (2nd derivative is negative). I did this by differentiating y=e^f(x) twice, but it is not always the case that if f(x) is concave down y=e^f(x) will have inflexion points.

g(x)=e^f(x)

let h(x) = e^x

.'. g(x)=h(f(x))
g'(x)=f'(x)*h'(f(x))(chain rule)=f'(x)*h(f(x)) (last bit is only working in this case because e^x is its own derivative)
hence g'(x) = f'(x)*g(x)
And that should give you the basic shape of the graph.

g''(x) = f''(x)*h(f(x))+f'(x)²*h(f(x)) = f''(x)*g(x)+f'(x)²*g(x)

And that'll give you points of inflection.

 

[Jumbo Cactuar]

y=e^f(x)
y'=f'(x)e^f(x)
y''=f'(x)²e^f(x) + f''(x)e^f(x)
y'''=f'(x)³e^f(x) + f'(x)²f''(x)e^f(x) + f''(x)²e^f(x) + f'''(x)e^f(x)

conditions of inflection;
y'' = 0, y''' not equal 0

f'(x)² + f''(x) = 0
f'(x)² = -f''(x)

f'(x)³ + f'(x)²f''(x) + f''(x)² + f'''(x) not equal 0
f'(x)³ + f'''(x) not equal 0

 

[Slidey]

Remind me again how any of this is helpful?

 

[withoutaface]

Remind me again how any of this is helpful?

If you've got the first derivative you can find the turning points.

If you've got the second derivative you can find the points of inflection.

This helps you sketch the function.

 

[shafqat]

Most HSC ext2 graph sketching question, including the example you gave from 2003, should be doable without any calculus.

 

[Slidey]

Most HSC ext2 graph sketching question, including the example you gave from 2003, should be doable without any calculus.

Exactly my point.

 

[withoutaface]

Exactly my point.

I just wanted to practice my compound functions. LEAVE ME ALONE!

 

[underthebridge]

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.

In regards to the third derivative, is it possible to see the nature of the third derivative on a graph like say we know second derivative is concavity. If the third derivative is zero, is there a horizontal point of inflection (this is as an aside)

 

[underthebridge]

Most HSC ext2 graph sketching question, including the example you gave from 2003, should be doable without any calculus.

Yes see with my example from 2003, the answer shows two points of inflection on either side of the y axis and below the axis. When i was originally drawing it i didnt include those points so im trying to find a way to find out whether a graph will have inflection points or not as opposed to using calculator guess and check (i find it difficult to determine concavity by using a calculator and guessing points anyhow)

 

[no_arg]

e^f(x)

This is a common question and it is important to appreciate that there is no real way to gauge the concavity of the function e^(f(x)) just by considering the graph of f!

Consider for example

f(x)=2ln(x)
g(x)=ln(x)
h(x)=(1/2)ln(x)

These three graphs are essentially the same...they will look identical in a simple sketch.
Yet

e^f is always concave up
e^g has zero concavity everywhere and
e^h is always concave down!

(over the natural domain of course)

Consider a function that makes a smooth transition near x=a from 1.001ln(x) to .99ln(x). It will look like a ln curve but e^f will have a point of inflection at x=a!

I'm afraid that making concavity decisions re e^f based on graphical data
is little more than wishful thinking!


Hope this helps

 

[Slidey]

I just wanted to practice my compound functions. LEAVE ME ALONE!

I noticed.

But I was responding to multiple threads: 1) the supposition that you would be able to use calculus in a HSC graphs question (on a 4u paper), and 2) that when not given f(x) it is possible to judge a change in concavity via a formula or rule, both of which are wrong, as no arg pointed out.

I think the most important thing to note is that a change in concavity IS possible (but not certain), and vigilance should be exercised accordingly when graphing insofar as it is possible.

 

[who_loves_maths]

why are ppl deviating from the topic of this thread?
first of all, it's obvious that when not explicitly given f(x) one cannot makes a decision on where a concavity change may occur on the composite e^f(x) for the very simple reason that one cannot draw a sketch of the curve y=e^f(x) ! (because you don't know what f(x) is).

secondly, the person who started this topic simply wanted to know under what conditions might POIs occur on a graph of e^(f(x)). and the answer is that it MAY only occur over a restricted domain of 'x' for which the graph y=(f(x)) is concave down, OR, for values of 'x' for which the graph y=f(x) exhibit horizontal POIs.
and all this is true ONLY IF f'''(x) + f'(x)3 does NOT equal to 0.

 

[Slidey]

Really, I think the talk about bringing calculus into it is what is off topic (which is fine so long as it is recognised as being secondary to the purpose of the thread). If you'll read carefully, you'll note the author is actually asking the question with specific regard to HSC graphing where f(x) is not given (and where calculus is typically not allowed).

 

[thunderdax]

It shouldn't matter where the inflexion points are unless they ask for it specifically. Usually I just put them somewhere in between a max and a min...

 

[who_loves_maths]

Really, I think the talk about bringing calculus into it is what is off topic (which is fine so long as it is recognised as being secondary to the purpose of the thread). If you'll read carefully, you'll note the author is actually asking the question with specific regard to HSC graphing where f(x) is not given (and where calculus is typically not allowed).

if you take a look at the original post that started this thread you'll notice that the author maintained no mention of "calculus is not allowed" or "specific regard to HSC graphing where f(x) is not given". In fact, the author him/herself specifically mentioned that he/she used calculus originally to find that in order for a POI to exist then the graph y=f(x) must be concave down for that value of 'x'.
so no, calculus is not what is off topic here. and no, the author did not highlight or even write the words 'calculus is not allowed'.

in fact, here is what the author initially stated for your convenience:

I have a question in regards to curves of the form y=e^f(x) where you are just shown a graph of y=f(x) (eg. 2003 HSC). I have noticed that occasionally there are inflexion points on these curves and I am not sure how to find out whether the curve will have inflexion points or not. I have worked out that if there are inflexion points on y=e^f(x), they will only occur where y=f(x) is concave down (2nd derivative is negative). I did this by differentiating y=e^f(x) twice, but it is not always the case that if f(x) is concave down y=e^f(x) will have inflexion points.

see how the words "differentiating", "derivative", and "where you are just shown a graph of y=f(x)" appear in the original quote?