Math Help - Global Extrema Help

1. Global Extrema Help

I have already solved both of these questions. Mainly, I just need somebody to check my work and tell me if it's correct. I'd really appreciate it!

Problem 1:
Find all local and global minimums and maximums of
$f(t) = 4t3 - 3t2 - 90t+21$
in the interval $-4 \leq t \leq 6$. List the coordinates of each below. If an item doesn't exist, write DNE.

First I found the critical points and classified them as local max's or min's.

$f\prime(x) = 12t^2 - 6t -90$

$12t^2 - 6t-90 = 6(t-3)(2t + 5)$

$t = 3, -\frac{5}{2}$

Second Derivative Test

$f\prime\prime(x) = 24t - 6$

$f\prime\prime(3) = 24(3)-6 = 66$

$f\prime\prime(-\frac{5}{2}) = 24(-\frac{5}{2}) - 6 = -66$

So f(x) has a local max at $(-\frac{5}{2}, 164.75)$ and a local min at $(3, -168)$.

Then I tried to find the global extrema:

$f(-4) = 77$
$f(-\frac{5}{2}) = 164.75$
$f(3) = -168$
$f(6) = 237$

So I said that f(x) has a global min of -168 and no global max on this interval $( -4 \leq x \leq 6 )$

Is all of that correct?

Problem 2:
Find the exact global maximum and minimum values of the function
$f(x) = sin(x) - xcos(x) - \frac{\pi}{2}cos(x)$

First I tried to find the critical points:

$f\prime(x) = sin(x)(x + \frac{\pi}{2})$

$sin(x)(x + \frac{\pi}{2}) = 0$

$x = -\frac{\pi}{2}, 0 , \pi$

At this point I realized that since there's no interval given, you could have infinitely many critical points with the unit circle, so I decided to graph the function. This is the graph:

After seeing this, I wrote the following as my answer:

There is no global extrama because there is no interval and the graph keeps fluctuating bigger as x approaches negative and positive infinity.

Is that correct? Is that a good way of wording it?

Thank you so much to anybody that replies!

2. Originally Posted by lysserloo
I have already solved both of these questions. Mainly, I just need somebody to check my work and tell me if it's correct. I'd really appreciate it!

Problem 1:
Find all local and global minimums and maximums of
$f(t) = 4t3 - 3t2 - 90t+21$
in the interval $-4 \leq t \leq 6$. List the coordinates of each below. If an item doesn't exist, write DNE.

First I found the critical points and classified them as local max's or min's.

$f\prime(x) = 12t^2 - 6t -90$

$12t^2 - 6t-90 = 6(t-3)(2t + 5)$

$t = 3, -\frac{5}{2}$

Second Derivative Test

$f\prime\prime(x) = 24t - 6$

$f\prime\prime(3) = 24(3)-6 = 66$

$f\prime\prime(-\frac{5}{2}) = 24(-\frac{5}{2}) - 6 = -66$

So f(x) has a local max at $(-\frac{5}{2}, 164.75)$ and a local min at $(3, -168)$.

Then I tried to find the global extrema:

$f(-4) = 77$
$f(-\frac{5}{2}) = 164.75$
$f(3) = -168$
$f(6) = 237$

So I said that f(x) has a global min of -168 and no global max on this interval $( -4 \leq x \leq 6 )$

Is all of that correct?

Problem 2:
Find the exact global maximum and minimum values of the function
$f(x) = sin(x) - xcos(x) - \frac{\pi}{2}cos(x)$

First I tried to find the critical points:

$f\prime(x) = sin(x)(x + \frac{\pi}{2})$

$sin(x)(x + \frac{\pi}{2}) = 0$

$x = -\frac{\pi}{2}, 0 , \pi$

At this point I realized that since there's no interval given, you could have infinitely many critical points with the unit circle, so I decided to graph the function. This is the graph:

After seeing this, I wrote the following as my answer:

There is no global extrama because there is no interval and the graph keeps fluctuating bigger as x approaches negative and positive infinity.

Is that correct? Is that a good way of wording it?

Thank you so much to anybody that replies!
On your first question, your domain is bound. Why do you think there isn't a global maxima?

On your second question, though the amplitude grows with x. Doesn't the first cycle give you the smallest peak and smallest trough by comparison to all the rest? Yaah, you may not know where x will end, but don't you think the peaks and the troughs have their beginning?

3. On the first question, the global max would be at the positive endpoint, which is x= 6 (so the global max would be 237). However, when I looked at the graph, at this point the graph just keep increasing going toward positive infinity. Are you saying that because you're given an interval, whatever happens outside of it should be disregarded? Would the mean that 237 is the global max?

On the second question, it's asking for a global maximum and minimum, meaning over the whole graph (because no interval is given). If the wave keeps getting bigger and smaller, how can there be global extrema? There's local extrema, but not global.

Or am I wrong? Could you explain?

4. Originally Posted by lysserloo
On the first question, the global max would be at the positive endpoint, which is x= 6 (so the global max would be 237). However, when I looked at the graph, at this point the graph just keep increasing going toward positive infinity. Are you saying that because you're given an interval, whatever happens outside of it should be disregarded? Would the mean that 237 is the global max?

On the second question, it's asking for a global maximum and minimum, meaning over the whole graph (because no interval is given). If the wave keeps getting bigger and smaller, how can there be global extrema? There's local extrema, but not global.

Or am I wrong? Could you explain?

The answer to your first paragraph is "yes."

The answer to your second paragraph is as follows:

If we were to rewrite you function into $f(x)=\sin x - A\cos x - \pi \cos x$, where $A=x$, it should tell you that the amplitude, $A$ will only grow--not subside. The $f(x)$will grow as $x$gets farther from the origin. Graph this generously, and you will see.

Now, think. Your domain is the set $(-\infty,\infty)$. Within your domain, closest to (0,0)you have the smallest peak and trough. The first trough to the left of (0,0) is you smallest minima, and to the right of (0,0), you have the smallest maxima. They are the smallest kids in the whole universe, aren't they? You may not know who is the tallest kid in the whole universe, but don't you at least know who is the shortest kid in the whole universe. If the kid is the shortest (i.e. minima)in the whole universe, isn't he also the smallest in that neighbohood? Now in that same neighboorhood, there is the tallest kid (maxima) who is the shortest amongst all the giants.

5. Okay I understand the first question. Thank you!

As far as the second question goes, I think maybe I'm just confused as to why what you're saying wouldn't be considered local extrema. If we're talking about "in the neighborhood" of things, isn't that a local max and min?

6. Originally Posted by lysserloo
Okay I understand the first question. Thank you!

As far as the second question goes, I think maybe I'm just confused as to why what you're saying wouldn't be considered local extrema. If we're talking about "in the neighborhood" of things, isn't that a local max and min?
Yes.

7. Originally Posted by novice
Yes.
Pardon me, the local max and min are to the left of (0,0). My graph shows me the Min. stands between -7rad and -6 rad, and Max. between -9rad and -10 rad. That's my graph; I did it in radians.

8. Okay, so if what you've just told me are indeed a local max and min, how does that answer my question, which is asking for global extrema?

I already knew there were local extrema, but over the course of the entire graph, on the interval $(-\infty , \infty)$, the max and min continue to get bigger and smaller, respectively. So...knowing that, how can you have a single global max (if it's continuously growing), or a single global min?

I'm really confused.

9. Originally Posted by lysserloo
Okay, so if what you've just told me are indeed a local max and min, how does that answer my question, which is asking for global extrema?

I already knew there were local extrema, but over the course of the entire graph, on the interval $(-\infty , \infty)$, the max and min continue to get bigger and smaller, respectively. So...knowing that, how can you have a single global max (if it's continuously growing), or a single global min?

I'm really confused.
You are not confused at all. You are smart. You answered your own question. You may close this thread now.