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

1. Sep 15, 2013

### s3a

1. The problem statement, all variables and given/known data
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.)

2. Relevant equations
Differentiating (I think).

3. 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!

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2. Sep 15, 2013

### Ray Vickson

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?

3. Sep 15, 2013

### CAF123

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)?

4. Sep 15, 2013

### pasmith

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 $(t + 5)(t + 7) < 0$.

5. Sep 15, 2013

### Ray Vickson

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.

6. Sep 16, 2013

### 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)?

7. Sep 16, 2013

### pasmith

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 $f(-5)$?

8. Sep 16, 2013

### Ray Vickson

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

9. Sep 16, 2013

### s3a

Edit: Sorry, I double posted.

10. Sep 16, 2013

### 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?

11. Sep 16, 2013

### MrAnchovy

Look at that again.

12. Sep 16, 2013

### Ray Vickson

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

13. Sep 16, 2013

### Dick

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.

14. Sep 16, 2013

### 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)?

15. Sep 16, 2013

### 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. :)

16. Sep 16, 2013

### Dick

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.

17. Sep 16, 2013

### s3a

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?

18. Sep 16, 2013

### Dick

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.

19. Sep 17, 2013

### 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?

20. Sep 17, 2013

### 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?