1. Let f(t) = (e^t - e^-t) / (e^t + e^-t). This isn't a logistic function, but becomes one if you move the t-axis, L / (1 + C e^-kt).
To begin with, what does that really mean? Shifting the t axis up or down means adding a constant, call it D. So for some D, f(t) + D = L/(1+Ce^-kt).
a) Graph this f.
Without the axes, this is pretty easy - we're told it's a logistic function! If we look at t very positive, the e^-t's die and the e^t's dominate, and it goes to 1. If we look at t very negative, the e^t's die and the e^-t's dominate, so it goes to -1. We did that sort of thing over and over with the logistic function.
So far we know the asymptotes, f=1 and f=-1. But logistic functions are supposed to bottom out at 0, not -1; so we're going to have to shift it up by 1 to make it look logistic.
When we do, we get f(t)+D between 0 and 2. That tells us L=2, since logistic functions go take values between 0 and L.
When does this f reach its inflection point, that halfway point, f=0? That meas e^t = e^-t, so e^2t = 1, so 2t = log 1 = 0, so t=0.
So far so good; we're looking for a logistic function 2/(1+Ce^-kt) that reaches halfway up (=1) at t=0. That means 1+Ce^-kt=2, or C=1.
For the last, we need to figure out k. There are two ways to do this, with or without calculus. Let's start without; we know that
(e^t - e^-t) / (e^t + e^-t) = f(t) - D = L/(1+Ce^-kt) - 1 = 2/(1 + e^-kt) - 1 = (1 - e^-kt)/(1 + e^-kt).
Multiply the guy on the right top & bottom by e^t, and if k=2 it turns into the guy on the left. Ta-da, k=2.
The calculus version: we know the derivative of a logistic function at its inflection point - kL/4. So let's evaluate the derivative of this guy at its inflection point. The derivative is
( (e^t+e^-t)^2 - (e^t-e^-t)^2) / (e^t+e^-t)^2
by the quotient rule. At t=0, which is where we know the halfway point to be, this is (4-0)/4 = 1. Since it's supposed to be kL/4 and we know L=2, we learn k=2.
In all: C=1, k=2, L=2.
2. Cost = $5/chair. Quantity q = 50-2p.
a. Costs = 5q = 5(50-2p), Revenue = pq = p(50-2p).
b. Profit = Revenue-Costs. The question asks to maximize profit. (If you insisted on maximizing revenue, you'll get it pointed out later!) Take the derivative as usual, and get p = 15.
c. You know the definition of Elasticity; plugging in p=15 you get E=3/2.
d. "Isn't that supposed to be 1? Why isn't it?" It's only supposed to be 1 if you're maximizing _revenue_. Since we were maximizing _profit_, it wasn't necessarily supposed to be 1, and indeed it wasn't.
3a. Draw a graph that has three critical points: from left to right, a local minimum, local maximum, then a critical point that is neither.
b. The sign of f', at the right, has to be negative. It's negative just to the right of the local maximum (since the function has to go down from there), and it doesn't start increasing again, or else the third critical point would be a local min.
4. f(x) = sin x/x (x nonzero), f(0)=1. Where are the local/global max/mins?
Look at the graph of sin x (inside the front cover of the book; also on the front page of the test, but anyway you should know this graph). For x>0 it's less than x, and concave down. The reverse are true for x<0.
So sin x/x < 1 for positive or negative x, which says that the value at x=0 is the absolute maximum (and thus a local maximum). Also sin x has the same sign as x in our interval [-1,pi] except at pi, so the quotient is nonnegative; therefore the zero at pi is the global minimum.
The local maxima and minima are a bit harder to account for (since the book doesn't have a picture of the tangent function - sigh). The derivative is (x cos x - sin x)/x^2 = (x - tan x)/(x^2 cos x). In the range [-1,pi] that interests us, x and tan x only touch each other at x=0, so the only places we can have local max/min are at 0 and the endpoints. Also, the endpoints _must_ be local maxima or minima since the derivative isn't zero there.
Since 0 is a max, the function is decreasing to each side, so the two endpoints have to be local minima. We already figured out that the one on the right is the global minimum.
In all: local maximum = global maximum at x=0, f(x)=1. local maxima at x=-1, f(x) = -sin(-1) = sin(1), and x=pi, f(x)=0. The one on the right is the global minimum.