F(t) = some_integral; where does the maximum occur?

[s3a]

Homework Statement

Given f(t) = integral {(x^2 + 12x + 35)/(1 + cos^2 (x)) dx}.

At what value of t does the local max f(t) occur?

(A more aesthetically-pleasing version of the question is attached as TheQuestion.jpg.)

Homework Equations

Differentiating (I think).

The Attempt at a Solution

I know I can get f'(t) by just replacing the “x”s with “t”s and removing the integral because of the way the limits of integration are set up and, differentiating and integrating undo each other.

I also know that I can find minima and maxima using f'(t) = 0 but, that gives me two answers t = –7 or t = –5 and, I need to know which one it is.

Am I expected to find f''(t) = 0 to see which is a minimum and which is a maximum? (I ask because that seems quite complex.)

Or, am I supposed to just pick values of t slightly larger and slightly smaller than each value of t that I have to determine this? (This doesn't sit well with me because, while it might work well in practice a lot of the time, it isn't theoretically fool-proof.)

Any input would be greatly appreciated!

 

[Ray Vickson]

Homework Statement

Given f(t) = integral {(x^2 + 12x + 35)/(1 + cos^2 (x)) dx}.

At what value of t does the local max f(t) occur?

(A more aesthetically-pleasing version of the question is attached as TheQuestion.jpg.)

Homework Equations

Differentiating (I think).

The Attempt at a Solution

I know I can get f'(t) by just replacing the “x”s with “t”s and removing the integral because of the way the limits of integration are set up and, differentiating and integrating undo each other.

I also know that I can find minima and maxima using f'(t) = 0 but, that gives me two answers t = –7 or t = –5 and, I need to know which one it is.

Am I expected to find f''(t) = 0 to see which is a minimum and which is a maximum? (I ask because that seems quite complex.)

Or, am I supposed to just pick values of t slightly larger and slightly smaller than each value of t that I have to determine this? (This doesn't sit well with me because, while it might work well in practice a lot of the time, it isn't theoretically fool-proof.)

Any input would be greatly appreciated!

The question makes no sense at all. Your formula for f does not have a 't' in it anywhere, so how can it be a function of t?

 

[CAF123]

The question makes no sense at all. Your formula for f does not have a 't' in it anywhere, so how can it be a function of t?

I guess that the upper limit on the integral is t and the lower limit a constant.
@s3a: why do you think it is complicated to find f''(t)?

 

[pasmith]

Homework Statement

Given f(t) = integral {(x^2 + 12x + 35)/(1 + cos^2 (x)) dx}.

At what value of t does the local max f(t) occur?

(A more aesthetically-pleasing version of the question is attached as TheQuestion.jpg.)

Homework Equations

Differentiating (I think).

The Attempt at a Solution

I know I can get f'(t) by just replacing the “x”s with “t”s and removing the integral because of the way the limits of integration are set up and, differentiating and integrating undo each other.

I also know that I can find minima and maxima using f'(t) = 0 but, that gives me two answers t = –7 or t = –5 and, I need to know which one it is.

Am I expected to find f''(t) = 0 to see which is a minimum and which is a maximum? (I ask because that seems quite complex.)

Or, am I supposed to just pick values of t slightly larger and slightly smaller than each value of t that I have to determine this? (This doesn't sit well with me because, while it might work well in practice a lot of the time, it isn't theoretically fool-proof.)

Any input would be greatly appreciated!

Since $1 + \cos^2 t$ is strictly positive everywhere, f(t) is increasing when $t^2 + 12t + 35 = (t + 5)(t + 7) > 0$ and decreasing when $(t + 5)(t + 7) < 0$.

 

[Ray Vickson]

I guess that the upper limit on the integral is t and the lower limit a constant.
@s3a: why do you think it is complicated to find f''(t)?

Of course, I *could* guess that, but should not need to. Besides, maybe that guess is wrong. It is the responsibility of the OP to supply the correct answer.

 

[s3a]

Oops! The interval is from 0 to t!

Pasmith, thanks, I see that now but, I'm not too sure about how to find where the local maximum is located (other than computing f''(t) which seems really complicated).

CAF123, is there some trick to exploit (because, when I put this into software, I get some gigantic expression)?

 

[pasmith]

Oops! The interval is from 0 to t!

Pasmith, thanks, I see that now but, I'm not too sure about how to find where the local maximum is located (other than computing f''(t) which seems really complicated).

f is increasing when $(t + 5)(t + 7)$ is positive, ie when $t \geq -5$ and $t \geq -7$ (so that $t \geq -5$) or when $t \leq -5$ and $t \leq -7$ (so that $t \leq -7$) and decreasing when $(t + 5)(t + 7)$ is negative, ie. when $t \geq -5$ and $t \leq -7$ (which is impossible) or when $t \leq -5$ and $t \geq -7$ (so that $-7 \leq t \leq -5$).

Thus f is increasing when $t \leq -7$, decreasing when $-7 \leq t \leq -5$, and increasing again when $t \geq -5$. What does that tell you about $f(-7)$ and $f(-5)$?

 

[Ray Vickson]

Oops! The interval is from 0 to t!

Pasmith, thanks, I see that now but, I'm not too sure about how to find where the local maximum is located (other than computing f''(t) which seems really complicated).

CAF123, is there some trick to exploit (because, when I put this into software, I get some gigantic expression)?

Well, computing f''(t) is likely the easiest way to go: just evaluate f''(-5) and f''(-7).

 

[s3a]

Edit: Sorry, I double posted.

 

[s3a]

Ray Vickson, are you referring to the derivative of (x^2 + 12x + 35)/(1 + cos^2 (x)) or (x^2 + 12x + 35)?

Pasmith, what that tells us is that -5 and -7 are maximums and, since -5 is closer to 0, we choose -5 as the local maximum?

 

[pbuk]

Pasmith, what that tells us is that -5 and -7 are maximums...

Look at that again.

 

[Ray Vickson]

Ray Vickson, are you referring to the derivative of (x^2 + 12x + 35)/(1 + cos^2 (x)) or (x^2 + 12x + 35)?

Pasmith, what that tells us is that -5 and -7 are maximums and, since -5 is closer to 0, we choose -5 as the local maximum?

Please assume that I meant exactly what I wrote: $f''(t) = \frac{d}{dt} f'(t)$.

 

[Dick]

Please assume that I meant exactly what I wrote: $f''(t) = \frac{d}{dt} f'(t)$.

Just use pasmith's advice and sketch a graph of the curve using the information about where it is increasing and decreasing. You don't even need the second derivative.

 

[s3a]

Looking at it again (from the left toward the right), I see that f is increasing to a maximum then decreasing to a minimum then increasing again such that the only local (explicitly definable – read the next paragraph) maximum is at t = -7.

Looking at the graph of f using software, it seems that there is a “global maximum” at t “=” infinity. Would it actually be correct to state that there is a global maximum at f(t) as t approaches infinity (or, equivalently, that there is a global maximum at f(t) as t approaches negative infinity)?

 

[s3a]

Ray Vickson, it's been a long time since I used the quotient rule; I looked at the quotient rule on Wikipedia and realized that it is in fact reasonable to be able to do this. :)

 

[Dick]

Looking at it again (from the left toward the right), I see that f is increasing to a maximum then decreasing to a minimum then increasing again such that the only local (explicitly definable – read the next paragraph) maximum is at t = -7.

Looking at the graph of f using software, it seems that there is a “global maximum” at t “=” infinity. Would it actually be correct to state that there is a global maximum at f(t) as t approaches infinity (or, equivalently, that there is a global maximum at f(t) as t approaches negative infinity)?

Yes, the only local maximum is at t=(-7). If you look at your graph you should see f(t) approaches infinity as t->infinity and -infinity as t->(-infinity) but I wouldn't call them global maxes or mins. I would say there is no global max or min.

 

[s3a]

Yes, the only local maximum is at t=(-7). If you look at your graph you should see f(t) approaches infinity as t->infinity and -infinity as t->(-infinity) but I wouldn't call them global maxes or mins. I would say there is no global max or min.

What if there was a global maximum (there can be more than one if they have the exact same y value, right?)? How do I know what's local and what's not? I'm assuming that local would mean identifying an interval to examine but, why wasn't an interval given in this case? Was it simply because whoever wrote the problem knew that there was only one maximum so, he or she did not deem it necessary to mention an interval?

 

[Dick]

What if there was a global maximum (there can be more than one if they have the exact same y value, right?)? How do I know what's local and what's not? I'm assuming that local would mean identifying an interval to examine but, why wasn't an interval given in this case? Was it simply because whoever wrote the problem knew that there was only one maximum so, he or she did not deem it necessary to mention an interval?

The question only asks about local maximums, doesn't it? At least that's what you posted. Knowing whether it's global or not usually depends more information than just the derivatives. You have to add to that the behavior of the function at the endpoints of an interval or at infinity.

 

[s3a]

Yes, it does only ask for a local maximum but, how do I know what's local and what's not (in general)? What if there were infinite “hills” for this particular function f(t) when approaching infinity such that there was many options. How would I determine that the maximum at t = -7 was the local maximum? Would it be because the lower limit of the integral was 0 and t = -7 is the closest maximum to 0?

 

[s3a]

Yes, it does only ask for a local maximum but, how do I know what's local and what's not (in general)? What if there were infinite “hills” for this particular function f(t) when approaching infinity such that there was many options. How would I determine that the maximum at t = -7 was the local maximum? Would it be because the lower limit of the integral was 0 and t = -7 is the closest maximum to 0?

 

[Ray Vickson]

What if there was a global maximum (there can be more than one if they have the exact same y value, right?)? How do I know what's local and what's not? I'm assuming that local would mean identifying an interval to examine but, why wasn't an interval given in this case? Was it simply because whoever wrote the problem knew that there was only one maximum so, he or she did not deem it necessary to mention an interval?

The problem of global optimization--even for a univariate function on a known interval--is not easy unless the function has some special properties. On an interval, a function can have hundreds of local maxima and minima, and the only known way, in general, to know the global optimum is to compute all the locals and pick the best one. Some researchers have developed algorithms for numerical global optimization, and the field is still under development. Google 'univariate global optimization' to see what is happening recently. Such research aims at sequentially establishing simple upper and/or lower bounds on f(x) to isolate intervals in which the global optimum will be found. None of them is straightforward or quick or efficient.

 

[s3a]

Edit: Sorry, I double posted! Why does this keep happening to me? :@

 

[s3a]

Ray Vickson, while that is interesting, I wanted to know how I can determine which extremum is (the most) local and not how to find a global extremum efficiently. (For the purpose of this course, I am currently happy with the method of computing all local extrema until I find the largest one.)

So, could you (or anyone else, for that matter) please help me figure out how I can determine which extremum is (the most) local?

 

[pbuk]

You misunderstand the meaning of local maximum. Try Mathworld. To continue your 'hills' analogy, every hill is a local maximum - it doesn't need to be Mount Everest, or even the biggest hill you can see.

The expression "the most local extremum" has no meaning.

There are some implications of this (i) if a global maximum exists it is also a local maximum; (ii) there may be no local maximum over a given domain (including an infinite domain); (iii) there may be more than one local maximum over a given domain, but if this is the case there must be a local minimum between them (this is why I knew your answer that f'(x)=0 was true at exactly two points and they were both local maxima was wrong).

The question setter probably said "the local maximum" rather than "a local maximum" to give you a hint that there was only one, so when you found two solutions for f'(x)=0 you had more work to do.

 

[s3a]

You misunderstand the meaning of local maximum. Try Mathworld. To continue your 'hills' analogy, every hill is a local maximum - it doesn't need to be Mount Everest, or even the biggest hill you can see.

The expression "the most local extremum" has no meaning.

There are some implications of this (i) if a global maximum exists it is also a local maximum; (ii) there may be no local maximum over a given domain (including an infinite domain); (iii) there may be more than one local maximum over a given domain, but if this is the case there must be a local minimum between them (this is why I knew your answer that f'(x)=0 was true at exactly two points and they were both local maxima was wrong).

The question setter probably said "the local maximum" rather than "a local maximum" to give you a hint that there was only one, so when you found two solutions for f'(x)=0 you had more work to do.

Oh, I think I get it now! Thank you all very much! :D