Math 112, Calculus I Final Exam Solutions

Math 112, Calculus I
Final Exam Solutions
David Murphy
1. (30 points) Let f be the function given by f(x) = √ x + 1. The graph of f crosses the x-axis
at point P = (−1, 0) and the y-axis at point Q = (0, 1).
(a) Write an equation for the line passing through points P and Q.
To find the equation of the line passing through points P and Q, we need to know its
slope, m, and a point on the line, say Q = (0, 1). Now the slope of the line is equal to
the rise divided by the run between P and Q, so
m = y Q − y P
x Q − x P
=
Hence, the equation of the line through P and Q is
(1) − (0)
(0) − (−1) = 1 1 = 1.
y − y Q = m(x − x Q ) =⇒ y − 1 = [1](x − 0) = x or y = x + 1.
(b) Write an equation for the line tangent to the graph of f at point Q. Show your work.
Again, to write the equation of the line tangent to the graph of y = f(x) = √ x + 1
at Q, we need to know the slope of the line and a point on the line. For the point
on the line, we’ll use Q, and to calculate the slope, we evaluate the derivative, which
is the slope of the tangent line after all, when x = 0 (the x-coordinate of Q). Now
f ′ (x) = d
dx [(x + 1)1/2 ] = (1/2)(x + 1) −1/2 [1] = 1
2 √ x+1 , so f ′ 1
(0) = √ = 1 2 (0)+1 2
is the
slope of the tangent line. Hence the equation of the tangent line is
y − y Q = m(x − x Q ) =⇒ y − 1 = 1 2 (x − 0) or y = 1 2 x + 1.
(c) Find the x-coordinate of the point on the graph of f, between points P and Q, at which
the line tangent to the graph of f is parallel to the line P Q.
The graph of y = f(x) = √ x + 1 is parallel to the the line P Q at the value of x
where the derivative, f ′ (x) = 1 2 (x + 1)−1/2 is equal to the slope of P Q, which is 1 as
computed in part (a) above. Hence, we want to solve the equation 1 2 (x + 1)−1/2 = 1 for
x. Multiplying both sides by (x + 1) 1/2 , we find that 1/2 = √ x + 1. Squaring both sides
gives 1/4 = x + 1, so that
x = 1 4 − 1 = −3 4
is the x-coordinate of the point on the graph of f between points P and Q where the
tangent line is parallel to P Q.
1

Math 112, Calculus I
Final Exam Solutions
2. (20 points) Suppose that a piston is moving straight up and down and that its position at
time t seconds is
s(t) = A cos(2πbt),
with A and b positive. The value of A is the amplitude of the motion, and b is the frequency
(number of times the piston moves up and down each second). What effect does doubling
the frequency have on the piston’s velocity, acceleration, and jerk (the instantaneous change
in acceleration)?
With b fixed, we have the following formulas for v(t), a(t), j(t):
v(t) = s ′ (t) = d [A cos(2πbt)] = A · [− sin(2πbt) · (2πb)] = −2πbA sin(2πbt),
dt
a(t) = v ′ (t) = d dt [−2πbA sin(2πbt)] = −2πbA[cos(2πbt) · (2πb)] = −4π2 b 2 A cos(2πbt),
j(t) = a ′ (t) = d dt [−4π2 b 2 A cos(2πbt)] = −4π 2 b 2 A[− sin(2πbt) · (2πb)] = 8π 3 b 3 A sin(2πbt).
Now, if we replace b with 2b, then we have the following
so that
s(t) = A cos(2π(2b)t) = A cos(4πbt)
v(t) = s ′ (t) = d [A cos(4πbt)] = A[− sin(4πbt) · (4πb)] = −4πbA sin(4πbt),
dt
a(t) = v ′ (t) = d dt [−4πbA sin(4πbt)] = −4πbA cos(4πbt) · (4πb) = −16π2 b 2 A cos(4πbt),
j(t) = a ′ (t) = d dt [−16π2 b 2 A cos(4πbt)] = −16π 2 b 2 A[− sin(4πbt) · (4πb)] = 64π 3 b 3 A sin(4πbt).
Thus, comparing corresponding amplitudes, we see that the amplitude for the velocity is
doubled, the amplitude of the acceleration is quadrupled (i.e., multiplied by 4), and the jerk’s
amplitude is 8 times that for the corresponding amplitudes of the velocity, acceleration, and
jerk before doubling the frequency. (So you can see why machines break down sooner when
we run them faster!)
3. (20 points) Use the limit definition of the derivative and the limit laws to find f ′ (2) if
f(x) = 1 − x . Write with complete detail.
2x
1−(2+h)
f ′ f(2 + h) − f(2)
2(2+h)
− 1−(2)
−1−h
2(2)
4+2h
(2) = lim h→0 = lim h→0 = lim − −1
4
h→0
h
h
h
4(−1−h)−(−1)(4+2h)
4(4+2h)
= lim h→0
h
= lim h→0
−4 − 4h + 4 + 2h
4(4 + 2h)h
= lim h→0
−2
4(4 + 2h) = −2
4[4 + 2(0)] = −2
16 = −1
8 .
= lim h→0
−2h
4(4 + 2h)h

Math 112, Calculus I
Final Exam Solutions
4. (20 points) Evaluate each limit, if it exists, or explain why it does not exist. If you use
L’Hôpital’s rule, state why it is applicable.
(a)
x 2 − 25
lim
x→−5 x + 5
= lim (x − 5)(x + 5)
x→−5
= lim x→−5 (x − 5) = (−5) − 5 = −10
x + 5
√ √ a + h − a
(b) lim
= 1
h→0 h 2 √ because this is the limit definition of the derivative of f(x) =
a
√ x at x = a, and we know that f ′ (x) = 1 2 x−1/2 , so f ′ (a) = 1
2 √ a .
ln(1 + x) − x
(c) lim
x→0 x 2 is an indeterminate form of type 0/0, so we may apply L’Hôpital’s Rule
1
1+x
and instead consider lim · 1 − 1
x→0 , which is again of the form 0/0, so we apply
(d)
2x
−(1 + x) −2 [1]
L’Hôpital’s Rule a second time and look at lim x→0 =
2
ln(1 + x) − x
−1/2. Hence, by L’Hôpital’s Rule, the original limit lim x→0
x 2
−[1 + (0)]−2
2
= −1/2.
lim is an indeterminate form of type 0 0 , so we can’t use L’Hôpital’s Rule immediately
x→0 +xx
but must first transform this into either a 0/0 or ∞/∞ form. First, let y = x x and
consider ln y = ln(x x ) = x ln x. Then lim x→0 + x ln x is a 0 · −∞ form, which still
isn’t 0/0 or ∞/∞, but we may rewrite it as x ln x = ln x
1/x , so that lim ln x
x→0 + 1/x is
now at last of the form −∞/∞ and we may apply L’Hôpital’s Rule to consider instead
1/x
lim x→0 +
−1/x 2 = lim x→0 + −x = 0. Hence, by L’Hôpital’s Rule, lim x→0 + x ln x = 0 as
well. Therefore, as x ln x = ln(x x ) and we wanted to find lim x→0 + x x , we exponentiate
to recover x x and thus also the limit, so that
lim
x→0 xx = e 0 = 1.
+
=
5. (20 points) Consider the piecewise defined function
⎧
1
⎪⎨ x < 0
x 2 −1
f(x) = x
⎪⎩
2 0 ≤ x < 1
x 1 ≤ x
(a) Where does f have discontinuities? Classify them as removable, jump or infinite.
On the interval (−∞, 0), f(x) = 1 , which is a rational function and so is continuous
x 2 −1
(and differentiable) as long as it is defined. However, it is not defined, and so not
continuous when x = −1 because we’d be dividing by zero here. Thus f(x) is not
continous at x = −1 having an infinite discontinuity.
On the interval (0, 1), f(x) = x 2 is a polynomial and so it is continuous (and differentiable).
On the interval (1, ∞), f(x) = x is again continuous (and differentiable) because
it is a polynomial function.
At x = 0, where the definition of f changes from 1
x 2 −1 to x2 , we have
1
lim f(x) = lim
x→0− x→0 x 2 − 1 = 1
(0) 2 − 1 = −1

Math 112, Calculus I
Final Exam Solutions
while
lim f(x) = lim
x→0 + x→0 x2 = (0) 2 = 0.
+
As these two limits are not equal, the limit lim x→0 f(x) does not exist so f(x) is not
continous at x = 0 having a jump discontinuity.
At x = 1, the definition of f changes from x 2 to x, and we have
lim f(x) = lim
x→1− x→1 x2 = 1
−
and
lim f(x) = lim x = 1.
x→1 + x→1 +
Since these two one-sided limits both exist, agree, and are equal to f(1) = 1, f is
continuous at x = 1.
(b) Where is f not differentiable? Support your answer.
First of all, f is not differentiable at x = −1 nor at x = 0 since f is not continuous at
these points. We have already noted that f is differentiable on (−∞, −1), (−1, 0), (0, 1)
and (1, ∞), so the only other point to check is at x = 1. From the left, lim x→1 − f ′ (x) =
lim x→1 − 2x = 2 and from the right lim x→1 + f ′ (x) = lim x→1 + 1 = 1. Since these limits
are different, there is a corner when x = 1 so the function f is not differentiable at x = 1.
6. (20 points) Let f be the function defined by f(x) = (x 2 − 3)e x for all real numbers x.
(a) Find the vertical and horizontal asymptotes, if they exist.
Since f(x) is defined for all real numbers x, so that we’re never dividing by 0, there are
no vertical asymptotes. To decide if there are horizontal asymptotes, we consider
lim f(x) = lim
x→∞ x→∞ (x2 − 3)e x = ∞ · ∞ = ∞
so there is no horizontal asymptote as x → +∞. Yet,
lim f(x) = lim
x→−∞ x→−∞ (x2 − 3)e x = ∞ · 0
is indeterminate, so we’ll use L’Hôpital’s Rule by first rewriting (x 2 − 3)e x = x2 − 3
e −x ,
2x
which is of the form ∞/∞ as x → −∞. Thus, consider lim x→−∞ , which is again
−e−x 2
∞/∞, so compute lim x→−∞
e −x = 0 since e−x → ∞ while 2 is constant. Hence,
applying L’Hôpital’s Rule, we have lim x→−∞ (x 2 − 3)e x = 0, so that the line y = 0 is a
horizontal asymptote of f.
(b) Find the intervals of increase or decrease.
First, f ′ (x) = (x 2 − 3)[e x ] + (e x )[2x] = e x [x 2 − 3 + 2x] = e x (x + 3)(x − 1), which is always
defined. Thus the only critical numbers are when f ′ (x) = 0, so are x = −3, +1. Dividing
the real number line at these points, we check whether f is increasing or decreasing on
the intervals (−∞, −3), (−3, 1), (1, ∞) by evaluating
f ′ (−4) = e −4 (−1)(−5) = 5e −4 > 0 =⇒ f is increasing on (−∞, −3);
f ′ (0) = e 0 (−3)(+1) = −3 < 0 =⇒ f is decreasing on (−3, 1);
f ′ (2) = e 2 (+5)(+1) = 5e 2 > 0 =⇒ f is increasing on (1, ∞).

Math 112, Calculus I
Final Exam Solutions
(c) Find the local maximum and minimum values.
By the First Derivative Test, since f changes from increasing to decreasing when x = −3,
f has a local maximum at the point (−3, f(−3)) = (−3, [(−3) 2 − 3]e −3 ) = (−3, 6e −3 ).
Hence the local maximum value of f is 6e −3 . The First Derivative Test also tells us that
f has a local minimum at the point (1, f(1)) = (1, [(1) 2 − 3]e 1 ) = (1, −2e) since the f
changes from decreasing to increasing at x = 1. Thus the local minimum value of f is
−2e.
(d) Find the intervals of concavity and the inflection points.
We have f ′′ (x) = (e x )[2x + 2] + (x 2 + 2x − 3)[e x ] = e x [x 2 + 4x − 1], which is defined for
all x, so we find when f ′′ (x) = 0 to locate where f may change in concavity. Since e x is
never zero, f ′′ (x) = 0 when x 2 + 4x − 1 = 0, which by the Quadratic Formula is when
x = −4 ± √ 4 2 − 4(1)(−1)
= −2 ± √ 5. We still have to show that concavity changes, so
2(1)
on the interval (−∞, −2 − √ 5), f ′′ (−5) = e −5 [(−5) 2 + 4(−5) − 1] = e −5 [25 − 20 − 1] =
4e −5 > 0 so f is concave up. On (−2 − √ 5, −2 + √ 5), which includes the point where
x = 0, we have f ′′ (0) = e 0 [(0) 2 + 4(0) − 1] = −1 < 0, so f is concave down. Finally,
on (−2 + √ 5, ∞), which includes x = 1, we have f ′′ (1) = e 1 [(1) 2 + 4(1) − 1] = 4e >
0 so f is concave up. Therefore, concavity changes when x = −2 − √ 5 and again
when x = −2 + √ 5, so f has inflection points at the points (−2 − √ 5, f(−2 − √ 5)) =
(−2 − √ 5, ((−2 − √ 5) 2 − 3)e −2−√5 ) = (−4.236, 0.216) and at (−2 + √ 5, f(−2 + √ 5)) =
(−2 + √ 5, ((−2 + √ 5) 2 − 3)e −2+√5 ) = (0.236, −3.728).
7. (20 points) Consider the function f(x) = √ 1 + x near x = 0.
(a) Find the linearization L(x) for f(x) near x = 0.
Since f ′ (x) = 1 2 (1 + x)−1/2 [1], we have
L(x) = f(0) + f ′ (0)[x − 0] = √ 1
1 + 0 + [
2 √ 1 + 0 ](x − 0) = 1 + 1 2 x.
(b) Use your linear approximation to estimate f(0.5). Is this approximation too large or too
small? Support your answer.
f(0.5) ≈ L(0.5) = 1+ 1 2
(0.5) = 1.25. To determine whether this approximation is too big
or too small, we check the concavity: f ′′ (x) = −1
4 (1+x)−3/2 [1] so that f ′′ (0) = −1/4 < 0
implies that f is concave down at x = 0 so the linear approximation L(0.5) = 1.25 is too
large.
(c) Show that the approximation of √ 1 + x by its linearization at the origin must improve
as x → 0 by showing that lim x→0 f(x)/L(x) = 1.
√ √
f(x) 1 + x 1 + (0)
lim
x→0 L(x) = lim
x→0 1 + 0.5x = 1 + 0.5(0) = 1 1 = 1.

Math 112, Calculus I
Final Exam Solutions
8. (25 points) A balloon is in the shape of a cylinder with hemispherical ends of the same radius
as that of the cylinder. The balloon is being inflated at the constant rate of 261π cubic
centimeters per minute. At the instant the radius of the cylinder is 3 centimeters, the volume
of the balloon is 144π cubic centimeters and the radius of the cylinder is increasing at the
rate of 2 centimeters per minute. (The volume of a cylinder with radius r and height h is
πr 2 h and the volume of a sphere with radius r is 4 3 πr3 .)
(a) At this instant, what is the height of the cylinder?
The total volume of the balloon is the volume of the cylinder plus the volumes of the
two hemispheres, which together make the volume of a single sphere, so
V = πr 2 h + 4 3 πr3 .
When V = 144π and r = 3, we have 144π = π(3) 2 h + 4 3 π(3)3 = 9πh + 36π, so 9πh =
144π − 36π = 108π. Thus h = 12 cm at this moment.
(b) At this instant, how fast is the height of the cylinder increasing?
Differentiating the volume equation above with respect to time t, we obtain
dV
dt = π[(r2 )[ dh
dt
Substituting in dV
dt
[2π(3)(12) + 4π(3) 2 ](2) = 9π dh
cm/min.
] + (h)[2r
dr
dt ]] + 4 3
= 261π, r = 3, h = 12 and dr
dt
dt
+ 216π, so 9π
dh
dt
dr dh
π[3r2 ] = πr2
dt dt + [2πrh + 4πr2 ] dr
dt .
= 2, we have 261π = π(3) 2 dh
dt +
dh
= 261π − 216π = 45π. Hence
dt = 5
9. (25 points) It’s 10:30 p.m. Your snowmobile is out of gas and you are 3 miles due south of
a major highway. The nearest gas station on the highway is 5 miles east of your position;
it closes at midnight. You can walk 5 miles per hour on roads, but only 3 miles per hour
through snowy fields. (Hint: time = distance/rate)
(a) What is the best route to the gas station?
(b) Can you make it there before the gas station closes?
You’re going to walk diagonally a distance, which we’ll call s, through the fields to the highway,
and then a distance x along the highway to the gas station. Then s is the hypotenuse of
the triangle in the figure you were given. Call the third side of this triangle z, so by the
Pythagorean Theorem we have s 2 = 3 2 + z 2 = 9 + z 2 . Moreover, z + x = 5 since the gas
station is 5 miles east of the point on the highway due north of your current location. Hence
x = 5 − z. The time it takes to walk s miles through the snowy fields is given by s/3, since
you have to walk s miles and you are walking at the rate 3 mph. The time to walk x miles
along the highway is x/5 since you walk 5 mph on the highway. Thus the total time for your
trip, which is what you want to minimize, is
t = s 3 + x 5 = √
9 + z 2
3
+ 5 − z
5 .
To minimize the function t(z) above, we find its derivative and set it equal to zero:
t ′ (z) = [ 1 2 (9 + z2 ) −1/2 [2z]]
3
+ −1
5 = 0.

Math 112, Calculus I
Final Exam Solutions
z
Adding 1/5 to both sides, we get
3 √ = 1 9+z 2 5 , so 5z 3 = √ 9 + z 2 . Hence 25z2
9
= 9 + z 2 , so
9 = 25z2
9
− z 2 = 25z2 −9z 2
9
= 16z2
9 . Therefore z2 = 81
16 , so z = ± 9 4
of which only z = 9/4
is reasonable as a solution.
consider t ′ (0) =
To confirm that this value of z corresponds to a minimum,
√ (0)
− 1
3 9+(0) 2 5 = −1/5 < 0 so that t is decreasing for z < 9/4 and f ′ (4) =
√ (4)
− 1
3 9+(4) 2 5 = 4
15 − 1 5 = 1
15
> 0 so that f is increasing for z > 9/4. Thus, z = 9/4 corresponds
to a minimum for t(z), so if you are in this situation you should walk so that you reach the
highway 9/4 = 2.25 miles east of your current position (so you have to walk an addition 2.75
miles east along the highway to the gas station). This is your optimal route.
To determine if you’ll reach the gas station before it closes or not, compute
√
9 + (9/4) 2
t(9/4) =
+ 5 − (9/4) = 1.25 + 0.55 = 1.8
3
5
so with your best route, it will take you 1.8 hours to reach the gas station from the point
where your snowmobile is out of gas. As the gas station closes an hour and a half after the
time your snowmobile runs out of gas, you don’t have enough time to get there, so you won’t
make it to the gas station before it closes (you’ll be 18 minutes too late!).