Test 4 Solutions

I.    Graph the function f(x)=x4-5x2+4.
To sketch a graph of f(x) showing all the information asked for, you needed to determine each of the following: Incorporating the above information, your graph should end up looking something like this:

graph

II.    For each of the following functions, determine:

a. The intervals on which the function is increasing/decreasing.
b. All local, endpoint and absolute extremes.
c. All points on inflection.
d. All asymptotes.
e. All vertical tangents and/or cusps.

Note: due to time constraints, I'm just going to type in the answers, rather than complete solutions. Please see me or email me with any questions or disagreements.

1. g(x)=sin x cos x, 0<x<Pi

a. g'(x)=(cos x)2 - (sin x)2. Therefore, g'(x)=0 when x=Pi/4 or 3Pi/4, g'(x)>0 when Pi/4<x<3Pi/4, and g'(x)<0 elsewhere. (As always, g is increasing iff g'>0, and decreasing iff g'<0.)

b. Based on a., g(Pi/4)=0.5 is a local maximum, and g(3Pi/4)=-0.5 is a local minimum. Endpoint extremes are g(0)=0 (local minimum) and g(Pi)=0 (local maximum). The absolute extremes, then, are (Pi/4, 0.5) and (3Pi/4, -0.5) (absolute maximum and absolute minimum, respectively).

c. g''(x)=-4sin x cos x (which is -4g(x), interestingly enough). Therefore g''(x)=0 when x=0, Pi/2 or Pi. Since an endpoint can't be a point of inflection (why?), (Pi/2, 0) is the only point of inflection.

d. No asymptotes.
e. No vertical tangents/cusps.

2. h(x) = x(x-1)-2
a. h'(x) = -(x+1)(x-1)-3. Therefore, h'(x)=0 iff x=-1, and h'(x) DNE when x=1. Checking intervals reveals that h'(x) is positive when -1 <x<1, and h'(x) is negative when x<-1  and when x>1.

b. From a. we see that h has a local minimum at x=-1. Therefore, (-1, -1/4) is a local minimum. This is also the absolute minimum, since h(x) has no other local extremes and h(x) -> 0 as x -> +oo or -oo (see below: horizontal asymptotes.)

c. h''(x) = 2(x+2)(x-1)-4. Therefore, h''(x)=0 iff x=-2. Since h''(x) also changes sign at x=-2, the point (-2, -2/9) is a point of inflection.

d. The vertical asymptote is x=1, since h(x) -> oo as x -> -1.
The horizontal asymptote is y=0, since h(x) -> o as |x| -> oo. This is true because h(x) is a rational function, and the degree of the denominator is greater than the degree of the numerator (2 and 1, respectively).

e. No vertical tangents or cusps.

3. j(x) = (x-4)2/3 - 2x/3
 
j'(x)  =  (2/3)(x-4)-1/3 - 2/3
= (2/3)((x-4)-1/3 - 1)
Therefore, j'(x) = 0 iff (x-4)-1/3 = 1; that is, iff x = 5. Testing intervals, we see that j'(x)<0 when x<4 and when x>5, and that j'(x)>0 when 4 < x < 5.

b. From a., we see that j(x) has a local extremes at x=4 and at x=5. Note that, even though j'(4) DNE, j(4) DOES exist, and so there can be (and is) a local extreme there. Thus, (5, -7/3) is a local maximum, and (4, -8/3) is a local minimum.

There are no absolute extremes, since j(x) -> oo as x -> -oo, and j(x) -> -oo as x -> oo. (This is true since (2/3)x is of higher degree than (x-4)2/3, and therefore it dominates j(x) as |x| -> oo. So, as |x| -> oo, j(x) will behave like the function (2/3)x.)

c. j''(x) = (-2/9)(x-4)-4/3. Therefore, j''(x) is never equal to zero, so there is no point of inflection. (Note: x=4 is not a point of inflection, since j''(x) DNE at x=4.)

d. No horizontal and/or vertical asymptotes.

e. There is a vertical cusp at x=4. This is because j'(x) -> -oo as x -> 4-, but j'(x) -> +oo as x -> 4+.

III.    Solve each of the following optimization problems.

1. Find the greatest possible value of xy, given that x and y are both positive and x+y=100.

We're trying to minimize xy, so we want to write it as a function of just one variable. Since x+y=100, we may write y=100-x. Therefore, xy=x(100-x), so the problem becomes: find the absolute maximum of the function f(x) = x(100-x).

To find the absolute maximum, we look at f'(x). In this case, f'(x)=100-2x, which is zero iff x=50. Therefore, (50, 2500) is a local maximum of f(x). Since f'(x)<0 for x<50 and f'(x)>0 for x>50, it's easy to see that (50,2500) is the absolute maximum of f(x). Therefore, the solution to the problem is 2500. (This maximum is achieved when x=50, y=50).

Note: in this problem, the information "x and y are both positive" isn't really necessary. However, it will be necessary in the following problem:

2. Find the greatest possible value of xy2, given that x and y are both positive and x+y = 100.
Proceeding as in problem #1, rewrite x as <>nobr>x = 100-y, and substitute. (Note that we could also use the substitution y = 100-x; however, the algebra is simplified by substituting for x instead.) Thus, xy2 becomes (100-y)y2, and so we are trying to maximize the function f(y) = (100-y)y2 = 100y2 - y3.

Note that f'(y) = 200y - 3y2 = 3y(200/3 - y), and so f'(y)=0 when y=0 and when y=200/3. Since f'(y)>0 when 0200/3, f(200/3)=(100/3)(200/3)2 is a local maximum.

Now, you need to be careful: f'(y)<0 when y<0; this implies that, as y -> -oo, f(y) -> oo. So, if y were allowed to be any real number, then f(y) would have no absolute maximum! Fortunately, though, we know that 0 < y < 100, since "x and y are both positive" (this was one of the conditions of the problem). Thus, the absolute maximum is either the local maximum or an endpoint maximum. Since f(y)=0 at both endpoints (y=0, y=100), we see, that f(y) has an absolute maximum at y=200/3.

Solution:
f(200/3)  =  (100-200/3)(200/3)2
= (100/3)(40000/9)
= 4000000/27 (or 148148.148..., 148 repeating)

3. Find the greatest possible value of x-xy, given that x1/2 + y1/2 = 1.
Rewrite y in terms of x, and then substitute:
y1/2  =  1 - x1/2
y  =  (1 - x1/2)2
 =  1 - 2x1/2 + x
Thus, the quantity we are concerned with, x-xy, can be rewritten as: x-x(1-2x1/2+x), which can be simplified:
x-xy  =  x - x(1 - 2x1/2 + x)
 =  x - (x - 2x3/2 + x2)
 =  2x3/2 - x2

Also, pay attention to the constraints on x implied here: clearly, x>0, since we're taking x1/2. But also, note that y1/2 = 1 - x1/2. Since we must have y>0 (for the same reason x>0; that is, we're taking its square root), we must then also have 1 - x1/2>0. Therefore, x1/2<1, and so x<1! (Similarly, y<1.) Combining these two conditions on x, we find that the domain of x is 0< x < 1.

If you don't quite understand the above paragraph, try the following experiment: choose a value for x greater than 1, such as 4 or 9. Then plug this into the original equation: x1/2 + y1/2 = 1, and try to solve for y1/2. You'll get a negative number, which is not possible (since 1/2-powers are positive by definition). This is why the equation can only be solved for values of x between 0 and 1.

Thus, our goal is to find the absolute maximum of the function f(x)=2x3/2 - x2, with domain 0<x<1. So, find the derivative: f'(x) = 3x1/2 - 2x. Setting f'(x)=0 and solving for x, we find that f'(x)=0 at x=0 and x=9/4. Of course, since we know 0<x<1, we can ignore x=9/4.

Since x=0 is an endpoint, the above calculation proves that f has no local extremes in the interval [0,1]. Therefore, we simply need to test the endpoints, and choose the maximum. Since f(0) = 0 and f(1) = 2-1 = 1, we conclude that the absolute maximum is f(1)=1.

Solution: the greatest possible value of x-xy is 1. (i.e., x=1, y=0)

Note: this is a good example of an optimization problem with a solution which is not a local extreme. Always check the endpoints, not just the points where the derivative is zero!

4. A rectangular fence is to be built such that its length plus twice its width totals 30 feet. What should the dimensions be in order to maximize the enclosed area?
Let L=length, W=width. Then, L+2W = 30 is the constraint on L and W. Since A=LW and L=30-2W, we can write A (area) as a function of W: A(W) = (30-2W)W = 30W - 2W2. The domain of this function is [0,15], since it's clear that W is neither negative nor greater than 15 feet. (Why not greater than 15 feet?)

Now, A'(W) = 30-4W, which is zero when W=7.5. This is a local maximum, so the absolute maximum occurs either at W=7.5 or at an endpoint (W=0 or W=15). Testing these three values, we get:

A(7.5) = (15)(7.5) = 112.5
A(0) = (30)(0) = 0
A(15) = (0)(15) = 0

Therefore, the maximum area is 112.5 sq. ft.

Solution: the fence's length should be 15 ft., and its width should be 7.5 ft.

5. A theatre is selling tickets to a play. The manager believes that the theatre will sell out (500 tickets) by charging $20/ticket, and that each $1 increase in the ticket price will result in 20 fewer tickets being sold. (For example: at $21/ticket, 480 tickets would be sold.) What price should be charged to maximize the revenues from the ticket sales, and how many tickets would be sold at this price?
Our unknown quantities here are: price per ticket, and number of tickets sold. So, let t = number of tickets sold, and p = price per ticket (in dollars).

Our goal is to maximize revenues -- that is, tickets sold times price per ticket, or tp. To do so, we need to write this quantity as a function of one variable. This requires us to find the relationship between p and t. We know two things about this relationship:

1. When p=20, t=500.
2. Whenever p increases by 1, t decreases by 20.

The second condition should look familiar; it describes a linear relationship between p and t. In particular, it gives you the slope of a linear equation: m = (change in t)/(change in p) = -20. Thus, there is an equation of the form: t = -20p + b, where b is some constant (the t-intercept) to be determined. Plugging in t=500 and p=20, and solving for b, we find that 500 = -400 + b, and so b=900. Thus, the equation relating t and b is: t = -20p + 900.

Now, let R stand for revenues, so R=tp. This can now be rewritten as a function of p: R(p) = (-20p+900)p = -20p2 + 900p. Note that p>20 -- since the theatre sells out at $20, there will be no reason to charge any less than $20. There is no maximum price allowed (though, of course, too high of a price will result in zero ticket sales).

R'(p) = -40p + 900, and so R'(p) = 0 when p = 900/40 = 22.5. This is a local maximum. Now, as was the case in #1, notice that R'(p)<0 for all p<22.5, R'(p)>0 for all p>22.5, and there are no other critical numbers for R'(p). Therefore, it's clear that p=22.5 is the absolute maximum of R(p). (That is, the left endpoint at p=20 isn't necessary to ensure an absolute maximum at p=22.5).

Therefore, the best price to charge is $22.50. The number of tickets sold will be t = -20(22.5) + 900 = 450.

Solution: to maximize revenues, sell 450 tickets at $22.50/ticket (Total revenues will be 450(22.5) = $10,125.00)

IV.    Prove each of the following statements.

1. Suppose f(x) is continuous on [0,1], f(0)=0 and f'(x)<1 for all x in the interval (0,1). Is it possible that f(1)=1? If so, simply write "yes." If not, then explain why.

Note: any time you're given two choices, and they are:
A. Just write one word.
B. Write a proof.

...there's about a 99% chance that choice A isn't going to be the right one. The correct answer here is that, by the Mean Value Theorem, f(1) cannot equal 1 under these conditions.

The proof: First, assume f(1) = 1. (Since we want to show that this, in fact, is not true, this assumption should lead us to a contradiction.)

We know that f(0)=0, and that f(x) satisfies the conditions of the Mean Value Theorem. (What are those? Make sure to find out, if you don't know.) Therefore, there must be some value of c in the interval (0,1) such that f'(c)=[(f(1)-f(0))/(1-0)]. But (f(1)-f(0))/(1-0) = (1-0)/(1-0) =1. Therefore, f'(c)=1 for some c in the interval (0,1). But this is a contradiction, since we know that f'(x)<1 for all x in the interval (0,1)! This means that the original assumption is untrue; therefore, f(1) is not equal to 1. QED

(Note: QED is an abbreviation of some Latin phrase which I forget, but which means "which was to be shown.")

(Note also: Remember, a good way to prove something cannot be true is to assume that it is true, and then to show that this assumption leads to a contradiction.)

2. The function f(x) = x3+4x-1 has a zero in the interval (0,1) (by the Intermediate Value Theorem). Show that f(x) has no other real zeros.
(Or equivalently: show that the graph of f(x) has only one x-intercept, corresponding to its single x-intercept.)

To show this, first look at the derivative of f(x): f'(x) = 3x2+4. Clearly, f'(x) is positive for all values of x. Therefore, f(x) is always increasing, and therefore the graph can only cross the x-axis at one point. This shows that f(x) has no other zeros.

The above isn't quite a formal proof, but it would have been enough to receive full (or close to full) credit. A thorough proof would include an argument like the following:

Let x0 stand for the zero of f(x) mentioned in the problem -- that is, we're given f(x0=0. Since f'(x) is always positive, we know that f' is increasing on the real line. This means (by the definition of "increasing") that, if we choose any x > x0, then f(x) > f(x0). On the other hand, if x < x0, then f(x) < f(x0). In either case, we have either f(x) > 0 or f(x) < 0 (since f(x0=0). Therefore, f(x) <> 0 whenever x <> x0. (Note: <> means "is not equal to.") QED

One could also prove #2 using contradiction: Let x0 be the zero of f(x) mentioned in the problem; that is, f(x0)=0. Assume there exists another number, x1, such that f(x1)=0. Then, by Rolle's Theorem, there must be some number c in the interval (x0, x1) such that f'(c)=0. But this contradicts something we already know: f'(x) is positive, for all real numbers! Therefore, our assumption is proven incorrect, and so there is no other zero of f(x). QED (again)

3. Suppose f and g are differentiable functions with the following properties: Prove that f2(x) - g2(x) = C (where C is some constant.)

To show that a function is constant, show that its derivative is always zero. In this case, then, we'll be done if we can show that the derivative (with respect to x) of f2(x) - g2(x) is zero.
d
dx		 [f2(x) - g2(x)]  =  2f(x)f'(x) - 2g(x)g'(x)  (by the Chain Rule)
 =  2f(x)g(x) - 2g(x)f(x)  (since f'=g and g'=f)
 =  0

Therefore, f2(x) - g2(x) is a constant function, independent of x. QED

Note: Two important functions which satisfy these conditions are the hyperbolic sine and hyperbolic cosine, abbreviated sinh(x) and cosh(x) (respectively). Their definitions are: sinh(x) = [ex - e-x]/2, and cosh(x) = [ex + e-x]/2. A simple calculation (made easier if you use the Difference of Squares property) shows that cosh2(x) - sinh2(x) = 1, for all values of x.

Next semester, you'll learn how to find the derivatives of functions like these. For now, you'll have to just take my word for it that sinh(x) is the derivative of cosh(x), and vice versa.
 
 

Extra Credit.    Each of the following problems is worth up to 5 points.

Note: The first person to turn in a correct solution to Extra Credit problem #2 (below) will receive 5 extra credit points. (No credit will be given for problem #1, since it is identical to Extra Credit problem #1 on Test 5, which will be posted Monday 5/3.)

1. Let L be a differentiable function such that L'(x)=1/x for x > 0, and such that L(1)=0. (Note: there was a typo here on the test - sorry!) Prove that for any two positive numbers a and b, L(ab) = L(a) + L(b).

2. Find the smallest positive integer which is 1 more than a multiple of 5, 2 more than a multiple of 6, and 3 more than a multiple of 7.

Back to the Math 85 page.
Last modified 4/30/99
ludwick@math.temple.edu