yes, I got the 3.24 and -3.24 and the 0---but that doesn't tell us when the herd hits the wall, so to speak. I guess I will just have to go the long way for now until I learn some calculus...if I live long enough for that.
I somewhat dislike this problem because it depends theoretically on calculus, which you do not yet know (although you seem close to ready for it). However, the calculus part has been given to you, and what's left is a straight-forward algebra problem. By the way, that is typical of problems requiring solution by calculus: you use calculus to reduce the problem to one soluble by algebra.
The function N(t) gives the number of deer as a function of t.
[MATH]N(t) = -\ t^4 + 21t^2 + 100 \text { if } t \ge 0.[/MATH]
Notice that I slightly changed the statement of the problem because the population at time 0 is clearly 100 so there is no need to restrict N(t) to t greater than zero.
You can use N(t) and algebra to answer certain questions exactly. For example, when will the herd go extinct? (That is certainly hitting the wall in one sense.)
[MATH]u = t^2 \implies N(t) = -\ u^2 + 21u + 100 = -\ ( u - 25)(u + 4).[/MATH]
Very straight-forward algebra, a u-substitution to reduce a quartic to a simple quadratic. Now when the herd goes extinct, the number of deer is zero.
[MATH]N(t) = 0 \iff u = 25 \text { or } u = -\ 4 \implies t = \pm \sqrt{25} = \pm 5 \text { or } t = \sqrt{-\ 4} = \pm 2i.[/MATH]
But two of those four numbers are not real, and one is negative. Consequently, only one, namely plus 5, can apply.
We can check that answer
[MATH]-\ 5^4 + 21(5^2) + 100 = -\ 625 + 21(25) + 100 = -\ 525 + 525 = 0.[/MATH]
And notice we should really say
[MATH]0 \le t \le 5 \implies N(t) = -\ t^4 + 21t^2 + 100.[/MATH]
But many interesting questions cannot be answered exactly by looking directly at N(t) although you can get approximate answers by looking at its graph. For example, you cannot tell exactly when the number of deer is at its maximum or when the deer stop increasing and start decreasing. It is exactly that sort of question that differential calculus answers. There is a different function that is closely related to N(t) called N's derivative or N's rate of change that deals with such questions.
How to find that related function is the business of differential calculus. But this problem
GIVES you that function, namely
[MATH]0 < t < 5 \implies N'(t) = R(t) = - \ 4t^3 + 42t.[/MATH]
Notice that this function is not definable at t = 0 and t = 5, which explains why the problem unnecessarily restricted t > 0 for N(t). Notice also the notation of N'(t), which indicates that R(t) is an entirely different function from N(t) but related to it, which the notation R(t) does not do.
R(t) basically tells you where N(t) is increasing and where it is decreasing and at what values of t N(t) is at a maximum or minimum. That is the
relevance of the rate of change. If R(t) is greater than zero, N(t) is increasing. If R(t) is less than 0, N(t) is decreasing. If N(t) is at a local maximum or local minimum in an open interval, R(t) is zero.
So the
exact question that you are asking (when does the deer heard start its descent to extinction) is when the herd stops increasing, which means when R(t) = 0. And that is solvable
exactly by simple algebra.
[MATH]0 < t < 5 \text { and } R(t) = - \ 4t^3 + 42t = 0 \implies[/MATH]
[MATH]0 < t < 5 \text { and } -\ t(4t^2 - 42) = 0 \implies 0 < t < 5 \text { and } t = 0, \text { or } t = \pm \sqrt{\dfrac{42}{4}} \implies[/MATH]
[MATH]]t = \sqrt{10.5}.[/MATH]
Now the number that you were wondering about relating to where the graph of R(t) was maximum is not about the number of deer at all because R(t) does not describe the number of deer. R(t) describes a rate. The number that you were wondering can be calculated by using calculus to find R'(t), the rate of change of the rate of change, which is also described as N''(t), the second derivative of N(t), and then by using algebra to find where R'(t) = 0. This is what I did in my first post. That was misguided of me because you really wanted to know when the number of deer stopped increasing and started descending into extinction, which is where R(t) = 0. But the number you estimated between 1 and 3 is where the the herd of deer is increasing most rapidly, a different question. I think you started to think (mistakenly) that R(t) described the number of deer.
You have all the algebra you need to answer exactly virtually everything you could possibly want to know about this deer herd. What you do not have is the calculus to let you derive the additional functions needed. The problem gave you one, but it did not bother to explain what that function really meant, which is why I do not like this problem.
