Math 2250 Written HW #11 Solutions
1. An airplane begins its descent toward the runway when it is 4 miles from the touchdown point
and at an altitude of 1 mile.
(a) Find a, b, c, d so that the cubic function f (x) = ax3 + bx2 + cx + d describes a smooth
glide path for the airplane as pictured below (Hint: you have 4 pieces of information
about the function f (x). If you translate these pieces of information into 4 equations,
you will be able to solve for the four unknowns a, b, c, and d.)
H-4,1L
1
æ
-4
-3
-2
-1
æ
Answer: Since f (x) to gives the altitude of the plane when it is x miles from the runway,
we’re given four pieces of information in the statement of the problem:
f (0) = 0
f (−4) = 1
f 0 (0) = 0
f 0 (−4) = 0
The first two just say that the plane goes through the points (0, 0) and (−4, 1). The
third says that the plane’s trajectory must level off to horizontal when it lands. The last
says that the plane was in level flight at the point (−4, 1), which is the point where it
started descending.
Now, let’s use these four pieces of information. From the first equation, we know that
f (0) = 0, so
f (0) = a(0)3 + b(0)2 + c(0) + d
0=0+0+0+d
0 = d.
So in fact the function f (x) = ax3 + bx2 + cx for some a, b, and c. Now, the second
equation tells us that f (−4) = 1, so
f (−4) = a(−4)3 + b(−4)2 + c(−4)
1 = −64a + 16b − 4c.
1
We’ll keep this equation in our back pocket for later use.
Now, the last two equations tell us about the derivative of f (x) = ax3 + bx2 + cx, so
let’s first compute the derivative:
f 0 (x) = 3ax2 + 2bx + c.
Since the third equation says that f 0 (0) = 0, we know that
f 0 (0) = 3a(0)2 + 2b(0) + c
0=0+0+c
0=c
So now we know that c = 0 and so f (x) = ax3 + bx2 and f 0 (x) = 3ax2 + 2bx. Also,
we can simplify the equation 1 = −64a + 16b − 4c to read 1 = −64a + 16b. Finally, the
fourth equation tells us that
f 0 (−4) = 3a(−4)2 + 2b(−4)
0 = 48a − 8b.
So now we have the system
0 = 48a − 8b
1 = −64a + 16b
which we want to solve for the two unknowns. From the first equation we know that
48a = 8b,
so after dividing both sides by 8 we have that
b = 6a.
We can substitute this into the other equation:
1 = −64a + 16b
1 = −64a + 16(6a)
1 = −64a + 96a
1 = 32a
1
a= .
32
Since b = 6a, this tells us that b =
6
32
=
f (x) =
is the function we’re looking for.
2
3
16 ,
and we can conclude that
1 3
3
x + x2
32
16
(b) Assume the plane follows the path you found in part (a). When is the plane descending
at the greatest rate?
Answer: The plane is descending at the greatest rate when the function f 0 (x) (which
gives the rate of descent) has its absolute minimum given the constraint −4 ≤ x ≤ 0.
Therefore, we should look for critical points of the function
f 0 (x) =
3 2
6
3
3
x + x = x2 + x.
32
16
32
8
As always when looking for critical points, we differentiate:
f 00 (x) =
6
3
3
3
x+ = x+ .
32
8
16
8
This function always exists, so the only critical points occur when it equals zero:
3
x+
16
3
0= x+
16
3
3
− = x
8
16
−2 = x.
f 00 (x) =
3
8
3
8
The only critical point of f 0 (x) is at x = −2, so we just have to evaluate f 0 (x) at x = −2
and at the endpoints x = −4 and x = 0: the smallest value that results will be the
absolute minimum. Of course, we already know that f 0 (0) = 0 and that f 0 (−4) = 0, so
we can just compute
f 0 (−2) =
3
3
12 6
3 6
3
(−2)2 + (−2) =
− = − =− .
32
8
32 8
8 8
8
Since this is smaller than 0, we conclude that f 0 (x) achieves its absolute minimum at
x = −2.
In other words, the plane is descending at the greatest rate when it is 2 miles from the
runway, where its altitude is
f (−2) =
1
3
1 3
1
(−2)3 + (−2)2 = − + = mile.
32
16
4 4
2
2. Determine whether the following statements are true or false. If the statement is true, explain
why. If it is false, give an example which shows that it is false (called a “counterexample”).
(a) The sum of two increasing functions is increasing.
Answer: True. Assuming f (x) and g(x) are differentiable functions which are both
increasing, then we know that f 0 (x) and g 0 (x) are both positive. But then if h(x) =
f (x) + g(x), it must be the case that
h0 (x) = f 0 (x) + g 0 (x)
is also positive, since the sum of two positive numbers is positive. Therefore, the function
h(x) is also increasing.
3
(b) The product of two increasing functions is increasing.
Answer: False. Consider the functions f (x) = ex and g(x) = ex − 10. Notice that both
f (x) and g(x) have ex as their derivative. Since ex > 0 for all x, this means that f (x)
and g(x) are always increasing, as we can see in the following graph:
30
20
10
-4
-2
2
4
-10
Then the product of f (x) and g(x) is the function
h(x) = f (x)g(x) = ex (ex − 10) = e2x − 10ex .
But now
h0 (x) = 2e2x − 10ex ,
which is not always positive. Indeed, for x < ln(5), h0 (x) < 0 (for example, h0 (0) = −8).
Therefore, the function h(x) is not always increasing, as we can see in the graph:
120
100
80
60
40
20
-4
-2
2
4
-20
(In fact, with slightly more work we could have come up with an example where both
functions are always increasing but the product is always decreasing. I won’t go through
the calculations, but consider
f (x) = g(x) = arctan(x) −
π
.
2
Then f (x) and g(x) are the same increasing function as we can see on the graph:
4
-2 Π
-Π
Π
2Π
-2
but the product f (x)g(x) = (arctan(x) − π/2)2 is always decreasing:
-2 Π
-Π
Π
2Π
Where do these examples come from? Well, if f (x) and g(x) are increasing and differentiable, then we know that f 0 (x) > 0 and that g 0 (x) > 0. Now, the product
h(x) = f (x)g(x) is increasing where its derivative is positive. But the derivative of
h(x) is
h0 (x) = f 0 (x)g(x) + f (x)g 0 (x)
by the product rule. We know f 0 (x) and g 0 (x) are positive, but if either f (x) or g(x) (or
both) is negative, then the whole expression can be negative. Then it’s just a matter of
coming up with such functions.)
3. Use your accumulated calculus skills to sketch the graph of the function
g(x) =
x
.
1 + x2
Be sure to label all intercepts, local minima, local maxima, inflection points, asymptotes,
absolute minima, absolute maxima, etc.
Answer: First, we can identify intercepts in the usual way. The y-intercept occurs when
x = 0:
0
g(0) =
=0
1 + 02
so the y-intercept is at (0, 0).
The x-intercepts occur when g(x) = 0, meaning that
x
= 0,
1 + x2
5
which only happens when x = 0, so the only x-intercept is the origin.
Next, we can see immediately that there are no vertical asymptotes since g(x) is never undefined. As for horizontal asymptotes, we just need to evaluate the limits
x
1 + x2
x
1/x2
= lim
·
x→+∞ 1 + x2 1/x2
1/x
= lim 1
x→+∞ 2 + 1
x
lim g(x) = lim
x→+∞
x→+∞
=0
and
x
x→−∞ 1 + x2
1/x2
x
·
= lim
x→−∞ 1 + x2 1/x2
1/x
= lim 1
x→−∞ 2 + 1
x
lim g(x) = lim
x→−∞
= 0.
Hence, the graph of g(x) has a horizontal asymptote at y = 0, which it approaches as it goes
off to infinity in either direction. Now, to find local minima and local maxima, we look for
critical points. The first step is to differentiate using the quotient rule:
(1 + x2 )(1) − x(2x)
(1 + x2 )2
1 + x2 − 2x2
g 0 (x) =
(1 + x2 )2
1 − x2
.
g 0 (x) =
(1 + x2 )2
g 0 (x) =
The denominator is never zero, so this function is always defined and the only critical points
will be when g 0 (x) = 0:
1 − x2
(1 + x2 )2
0 = 1 − x2
0=
x2 = 1
x = ±1.
6
So the only critical points of g(x) are at x = 1 and x = −1. We compute that
3
1 − (−2)2
=− <0
(1 + (−2)2 )2
25
1 − 02
g 0 (0) =
=1>0
(1 + 02 )2
3
1 − 22
=− <0
g 0 (2) =
2
2
(1 + 2 )
25
g 0 (−2) =
so the sign of the derivative is given in the following diagram:
-
+
-1
1
Therefore, g(x) is decreasing on (−∞, −1), increasing on (−1,
1), and decreasing on (1, +∞),
and we can see that g(x) has a local minimum at −1, − 12 and a local maximum at 1, 12 .
Moreover, these must be the absolute minimum and the absolute maximum since the function
approaches zero as it goes off to both the left and the right.
Turning to the question of inflection points, let’s take the second derivative:
g 00 (x) =
g 00 (x) =
g 00 (x) =
g 00 (x) =
g 00 (x) =
g 00 (x) =
g 00 (x) =
g 00 (x) =
(1 + x2 )2 (−2x) − (1 − x2 ) · 2(1 + x2 ) · 2x
(1 + x2 )4
(1 + 2x2 + x4 )(−2x) − 4x(1 − x2 )(1 + x2 )
(1 + x2 )4
−2x − 4x3 − 2x5 − 4x(1 − x4 )
(1 + x2 )4
3
−2x − 4x − 2x5 − 4x + 4x5
(1 + x2 )4
2x5 − 4x3 − 6x
(1 + x2 )4
2x(x4 − 2x2 − 3)
(1 + x2 )4
2x(x2 − 3)(x2 + 1)
(1 + x2 )4
2x(x2 − 3)
,
(1 + x2 )3
where we arrive at the last line by canceling a factor of 1 + x2 from both the numerator and
the denominator.
Therefore, the possible inflection points occur when g 00 (x) = 0, meaning that
2x(x2 − 3)
(1 + x2 )3
0 = 2x(x2 − 3)
0=
7
√
so either x = 0 or x = ± 3. We can check whether these are really inflection points by
checking whether g 00 (x) switches sign. The signs of g 00 (x) are given by
-
+
-
- 3
+
0
3
as we can verify by evaluating g 00 (x) at the following points:
2(−2)((−2)2 − 3)
4
=−
<0
(1 + (−2)2 )3
125
1
2(−1)((−1)2 − 3)
= >0
g 00 (−1) =
2
3
(1 + (−1) )
2
2
2(1)((1) − 3)
1
g 00 (1) =
=− <0
2
3
(1 + (1) )
2
2
2(2)((2) − 3)
4
g 00 (2) =
=
>0
(1 + (2)2 )3
125
g 00 (−2) =
√
√
3),
concave
up
on
(−
3, 0), concave
Therefore, the function
g(x)
is
concave
down
on
(−∞,
−
√
√
down again on (0, 3), and concave up again on ( 3, +∞). Moreover, there are inflection
points at
√ !
√ !
√
√
3
3
− 3, −
, (0, 0),
3,
.
4
4
Putting this all together, then we can draw and label the graph of the function, which looks
like the following, with local extrema labeled with blue dots and inflection points with purple
squares:
1,
1
2
1
O
2
3,
æ
à
3
4
1
4
H0, 0Là
-1
-5
- 14
3
- 3,4
à
æ
- 12
1
-1, - O
2
8
1
5