Final Exam Review Sheet Solutions
1. Find the derivatives of the following functions:
Ÿ a) f HxL = ã2 x+3 × tanIx 3 M. f ' HxL = 2 ã2 x+3 × tanIx 3 M + ã2 x+3 × Isec2 Ix 3 M I3 x 2 MM. Product rule and chain rule used.
Ÿ b) gHxL = Ix 6 + 5 xM × ln4 HxL. g ' HxL = I6 x 5 + 5M × ln4 HxL + Ix 6 + 5 xM × I4 ln3 HxL H1 xLM. Product rule and chain rule
used.
2. Find the equation of the tangent line to the curve x y 2 + x 2 y 3 = 12 at the point H1, 2L.
We first find the derivative y ', where y = f HxL is the implicitly defined function of x whose graph is the "function-graph" portion
of the curve that passes through the point (1, 2). We find y ' by implicit differentiation:
y2 + 2 x y y ' + 2 x y3 + 3 x2 y2 y ' = 0
y ' I2 x y + 3 x2 y2 M = -y2 - 2 x y3
y' =
-y2 -2 x y3
2 x y+3 x2 y2
.
Therefore, the slope of the tangent line at (1, 2) is y '
given in point-slope form: y - 2 =
-5
4
x=1,y=2
Hx - 1L, or y = 2 -
5
4
=
-4-16
4+12
6
5
4
3
2
1
-3
-2
-1
1
-20
16
=
-5
.
4
The equation of the tangent line is easily
Hx - 1L. Here is a graph of the curve and the tangent line at (1, 2):
7
Out[12]=
=
2
3
2
final review answers.nb
3. Find the derivative of the function hHxL =
x 2 sinHxL
3
3
ãx +5
using logarithmic differentiation.
We begin by taking ln of both sides, and expanding on the RHS by using properties of
logs:
lnHhHxLL = ln
x2 sinHxL
3
3
ãx +5
= ln J
N
x2 sinHxL 13
3
ãx +5
=
1
3
lnJ
x2 sinHxL
3
ãx +5
N=
1
3
IlnIx2 sinHxLM - lnIãx + 5MM =
3
1
3
I2 lnHxL + lnHsinHxLL - lnIãx + 5MM .
3
Next, we differentiate both sides:
1
hHxL
h ' HxL =
1
3
Jx +
2
cosHxL
sinHxL
-
1
× ã x × 3 x2 N
3
3
ãx +5
Finally, we multiply by hHxL to solve for the desired h ' HxL:
h ' HxL = hHxL × J 3 J x +
1
2
cosHxL
sinHxL
-
1
3
ãx +5
× ãx × 3 x2 NN =
3
3
x2 sinHxL
3
ãx +5
×J 3 J x +
1
2
cosHxL
sinHxL
-
1
3
ãx +5
× ãx × 3 x2 NN.
3
4. Suppose that a sample of a radioactive element decays to 95% of its original amount after 2
years. Find the half-life of the element, and find the time required for the sample to decay to
10% of its original amount.
Let AHtL be the amont of the element remaining in the sample after t years have elapsed. Let A0 be the initial amount, that is, the
amount present at time t = 0. Then we know that AHtL = A0 ãk t , since the substance decays exponentially. To determine the value
of k, we plug in t = 2, when the amount will have decayed to 95% of its original value:
AH2L = A0 ãk 2 = .95 A0 Þ ã2 k = .95 Þ k =
In[13]:=
Out[13]=
k = [email protected]  2
lnH.95L
,
2
which is equal to
- 0.0256466
To find the half-life, we solve the equation AHtL = .5 A0 for t:
A0 ãk t = .5 A0 Þ ãk t = .5 Þ t =
In[14]:=
Out[14]=
lnH.5L
,
k
which equals
[email protected]  k
27.0268
The half-life is therefore approximately 27 years.
To find the time required for the sample to decay to 10% of its original size, we solve the equation AHtL = .10 × A0 :
A0 ãk t = .10 A0 Þ ãk t = .10 Þ t = ln (.10)/k, which has the value
In[15]:=
Out[15]=
[email protected]  k
89.7811
So it takes almost 90 years for the sample to decay to 10% of its original size. Here is a graph of the function AHtL and the
horizontal lines at 50% and 10% of the original amount, which is set to 100.
final review answers.nb
3
So it takes almost 90 years for the sample to decay to 10% of its original size. Here is a graph of the function AHtL and the
horizontal lines at 50% and 10% of the original amount, which is set to 100.
In[16]:=
PlotA9100 ãk t , 50, 10=, 8t, 0, 100<E
100
80
60
Out[16]=
40
20
20
40
60
80
100
5. The volume of a cube is increasing at a rate of 10 cm3 ‘ min. How fast is the surface area
increasing when the length of an edge is 30 cm?
Let the side of the cube be s. Then the volume is V HsL = s3 . We are given that d V  d t = V ' = 10 cm3 ‘ min. Therefore,
3 s2 s ' = 10, so s ' = 10 ‘ I3 s2 M cm min. The surface area of the cube is S = 6 s2 , since the cube has six square faces of side length
s. We want to find d S  d t = S ' when s = 30. We see that S ' = 12 s s ', and when s = 30 and s ' = 10 ‘ I3 × 302 M, we obtain
S ' = 12 × 30 × H10  H3 × 30 ^ 2LL = H12 × 10L  H3 × 30L = 12  9 = 4  3 cm2 ‘ min.
6. Find the linear approximation to f HxL = 25 - x 2 near x = 3. Use it to estimate f H3.08L. If
Annie were to approximate f H3.08L crudely by f H3.08L » f H3L = 4, what percentage error would
she make?
The linear approximation "formula" is f HxL » LHxL = f HaL + f ' HaL × Hx - aL, for x near to a. In our case, a = 3 and f HaL = f H3L = 4.
Moreover, f ' HxL =
of f H3.08L = 4 In[17]:=
Out[17]=
1
2
3
4
25-x2
H-2 xL, so f ' HaL = f ' H3L =
-3
=
25-32
-3
.
4
Therefore, LHxL = 4 -
3
4
Hx - 3L. Our approximate value
H3.08 - 3L = 4 - 3 H.02L = 4 - .06 = 3.94. Mathematica gives the "exact" value
25 - 3.08 2
3.93873
We see that 3.94 is an excellent approximation to the exact value.
Now suppose that Annie approximates f H3.08L by f H3L = 4. What is her percentage error? Using our approximation of 3.94 as
the exact value, we see that the absolute size of her error is .06, so her relative error is
In[18]:=
.06  3.94
Out[18]=
0.0152284
.06
3.94
expressed as a percent.
We see that Annie's relative error is about 1.5%. Notice that if we compare .06 to her approximate value of 4, we get essentially
the same answer for the percentage error.
4
final review answers.nb
We see that Annie's relative error is about 1.5%. Notice that if we compare .06 to her approximate value of 4, we get essentially
the same answer for the percentage error.
In[19]:=
.06  4
Out[19]=
0.015
7. Find the critical points of the function f HxL = x 4 Hx - 1L , and determine which correspond to
local extremes of f .
3
The function is defined and differentiable on H-¥, ¥L, so the only critical points are of the type x = c, where f ' HcL = 0. We have
that
f ' HxL = 4 x3 Hx - 1L + x4 × 3 Hx - 1L2 × 1 = x3 Hx - 1L H4 Hx - 1L + 3 xL = x3 Hx - 1L H7 x - 4L,
3
2
2
so the values of x at which f ' HxL = 0 are 0, 4  7, and 1. We use test values to determine the sign of f ' on the open intervals
determined by the critical points: On H-¥, 0L, we use x = -1, and find f ' H-1L > 0, so f is increasing on H-¥, 0L. On H0, 4  7L,
we use x = 2  7, and see that f ' H1  7L < 0, so f is decreasing on H0, 4  7L. This implies that x = 0 is a local maximum point. On
H4  7, 1L, we use x = 5  7, and find f ' H5  7L > 0, so f is increasing on H4  7, 1L, which implies that x = 4  7 is a local minimum
point. Finally, on H1, ¥L, we use x = 2, and find f ' H2L > 0, so f is increasing on H1, ¥L. Therefore, x = 1 is a "terrace point" of
the function f . Here is a graph confirming our analysis:
In[21]:=
Plot@x ^ 4 Hx - 1L ^ 3, 8x, - 1  2, 3  2<D
0.02
0.01
Out[21]=
-0.5
0.5
-0.01
-0.02
1.0
1.5
final review answers.nb
8. For the function f HxL = cos2 HxL - 2 sinHxL on the interval @0, 2 ΠD, find the sub-intervals on
which f is increasing and decreasing, and the sub-intervals on which f is concave up and
concave down. Also find the global maximum and minimum values of f on the interval.
f ' HxL = 2 cosHxL × H-sinHxLL - 2 cosHxL = -2 cosHxL × HsinHxL + 1L. Therefore, f ' HxL = 0 when cosHxL = 0 and when sinHxL = -1. On
the interval @0, 2 ΠD , cosHxL = 0 at x = Π  2 and x = 3 Π  2, and sinHxL = -1 at x = 3 Π  2. Therefore, we see that there are two
critical points on the interval: Π  2 and 3 Π  2. On the interval @0, Π  2L, we use test value x = Π  4 and find that f ' HΠ  4L < 0, so f
is decreasing on @0, Π  2L. On HΠ  2, 3 Π  2L, we use test value x = Π, and find f ' HΠL = 2 > 0, so f is increasing on HΠ  2, 3 Π  2L.
On H3 Π  2, 2 ΠD, we use test value x = 7 Π  4, and we find f ' H7 Π  4L = -2 H.707L H-.707 + 1L < 0, so f is decreasing on
H3 Π  2, 2 ΠD. This implies that x = Π  2 is a local minimum, and x = 3 Π  2 is a local maximum, of f . We can now determine the
global maximum and global minimum values by evaluating f at the critical points and the endpoints:
In[24]:=
f@x_D := Cos@xD ^ 2 - 2 Sin@xD
In[29]:=
Out[29]=
8f@0D, f@Π  2D, f@3 Π  2D, f@2 ΠD<
81, - 2, 2, 1<
From this we see that the global minimum (output) value is -2 and the global maximum value is +2, and that these occur at the
local minimum and local maximum points found earlier.
Finally, we study the concavity by investigating the second derivative: f '' HxL = H2 sinHxLL HsinHxL + 1L + H-2 cosHxLL HcosHxLL.
f '' HxL = 2 sin2 HxL + 2 sinHxL - 2 cos2 HxL = -2 Icos2 HxL - sin2 HxL - sinHxLM = -2 I1 - sin2 HxL - sin2 HxL - sinHxLM.
Therefore, the second derivative is 0 when the factor (-2 sin2 HxL - sinHxL + 1M = H-1L I2 sin2 HxL + sinHxL - 1M = 0, or
H2 sinHxL - 1L HsinHxL + 1L = 0
We see that the second derivative is 0 when sinHxL = 1  2, which happens at x = Π  6 and 5 Π  6 (on our interval), and when
sinHxL = -1, which occurs at x = 3 Π  2. By checking the sign of the second derivative, one checks that f is concave down on
@0, Π  6), concave up on HΠ  6, 5 Π  6L, and concave down on H5 Π  6, 2 ΠD. -- The concavity doesn't change at 3 Π  2. The following plot of the function f confirms this analysis.
NOTE: A problem such as this is definitely too hard to be included on the exam! However, the tools of the first and second
derivatives are used just as they would be in an easier case; one just has to work harder to handle the trig functions, etc.
5
6
final review answers.nb
In[23]:=
Plot@Cos@xD ^ 2 - 2 Sin@xD, 8x, 0, 2 Pi<D
2
1
Out[23]=
1
2
3
4
5
6
-1
-2
9. Find the following limits:
Ÿ a) limx®1
x 9 -1
x 5 -1
= limx®1
9 x8
5 x4
9
= .
5
L'Hospital's Rule used once; the original limit is of the form "0/0."
2
Ÿ b) limx®¥ x 3 ã-x = limx®¥
x3
ãx
2
=0
Either remember ãx dominates any power of x, or use L'Hospital's rule to reduce the power of x in the numerator until the
numerator is a constant.
Ÿ c) limx®0 H1 - xL1x :
Let the limit be L, and take logs to get lnHLL = limx®0 x HlnH1 - xLL = limx®0
1
1
Rule, we find that lnHLL = limx®0
In[31]:=
Out[31]=
N@1  ãD
0.367879
A plot confirms this result:
1-x
H-1L
1
lnH1-xL
,
x
which is of the form "0/0." Using L'Hospital's
= -1. Therefore, L = ãH-1L = 1  ã, which equals
final review answers.nb
In[30]:=
Plot@H1 - xL ^ H1  xL, 8x, - .2, .2<D
0.40
0.39
0.38
0.37
Out[30]=
0.36
0.35
0.34
-0.2
-0.1
0.1
0.2
10. A poster is to have an area of 180 in2 , with one-inch margins along the bottom and the
two sides, and a two-inch margin at the top. What dimensions will give the largest printed
area (the total area less the margins)?
Let w be the width and h the height of the overall poster. Then w× h = 180. The printed area will have a width of w - 2 and a
height of h - 3, to accommodate the margins as described in the problem. Therefore, the area of the printed region is
Hw - 2L Hh - 3L = w h - 2 h - 3 w + 6 = 180 - 2 h - 3 w + 6. From the constraint that w h = 180, or h = 180  w, we can write the
area of the printed region as a function of w: pHwL = 180 - 2 H180  wL - 3 w + 6. We need to find the global maximum of this
function on H0, ¥L. (Note that as w ® 0+ and w ® ¥, we have that pHwL ® -¥, so there will be a finite interval of "reasonable"
values of w for which the printed area is positive.)
y = pHwL
120
100
80
60
40
20
5
10
15
20
To find the maximum, we find the critical points of p: p ' HwL = 360 ‘ w2 - 3, and p ' HwL = 0 when 360 = 3 w2 , or w2 = 120, or
w=±
120 . The maximum we seek can only occur at the positive critical point w =
10.9545. So the dimensions of the optimal poster are w =
mately equal to 16.4317.
120 and h = 180 ’
120 =
120 , which is approximately equal to
180×180
120
=
270 , which is approxi-
7
8
final review answers.nb
11. If f ' HxL = 8 x 3 + 12 x + 3 and f H1L = 6, find f HxL.
The general antiderivative of 8 x3 + 12 x + 3 is 2 x4 + 6 x2 + 3 x + C. To determine the value of C, we use the initial condition
f H1L = 6 = 2 + 6 + 3 + C Þ C = -5. Therefore, f HxL = 2 x4 + 6 x2 + 3 x - 5.
12. The graph of f ' is shown. Sketch the graph of f , assuming that f H0L = -1.
y = f 'HxL and y = f HxL
2
-2
-1
1
2
3
4
5
-2
-4
13. A stone dropped off a cliff hits the ground with a speed of 120 ft/sec. What is the height
of the cliff? (Recall that the acceleration due to gravity is a constant aHtL = -32 (ft/sec)/sec
near the surface of the earth.)
Let us call the height of the cliff h. We know the initial position of the stone is sH0L = h, where the position is the height of the
stone above the canyon floor. We also know that the initial velocity of the stone is vH0L = 0, since the stone is "dropped" -- it is
not thrown up or down. Finally, we know that the acceleration is a constant: aHtL = -32 ft/sec2 . Taking the antiderivative of the
acceeleration to get the velocity, we find that vHtL = -32 t + C1 , and C1 = 0 since the initial velocity of the stone is 0. Taking the
antiderivative of velocity to get the position, we find that sHtL = -16 t2 + C2 . Plugging in t = 0, we find that
sH0L = h = -16 × 02 + C2 Þ C2 = h, so sHtL = -16 t2 + h. Now, at what time t is the instantaneous velocity equal to -120 ft/sec (this
will be the moment of impact)? Solve -32 t = -120 Þ t =
fore, we can now solve for h: 0 = sI
15
15 2
M = -16 I 4 M
4
120
32
=
15
sec.
4
15 2
+ h Þ h = 16 I
4
So the height of the stone is 0 at time t =
M = 225 ft. The cliff is 225 feet high.
15
4
sec. There-
14. Compute the left, right, and midpoint Riemann Sums of the function f HxL = x 2 - 6 x on the
interval @0, 6D using 3 subintervals. Then compute the same sums using 6 subintervals.
We first compute a table of values for the function f , that includes all of the values we will need to compute the requested
Riemann Sums:
final review answers.nb
x
1
0
fHxL 0 -
2
11
4
3
1
-5 -
2
27
4
5
2
-8 -
2
35
4
7
3
-9 -
2
35
4
4
-8 -
9
2
27
4
If we use three subintervals, the width of each subinterval is Dx =
5
11
-5 6-0
3
2
11
4
6
0
= 2. If we use a left sum, the three rectangle heights are
f H0L, f H2L, and f H4L, so the left sum is f H0L × Dx + f H2L × Dx + f H4L × Dx which is equal to 0*2 + (-8)*2 + (-8)*2 = -32. Here is a
picture:
LeftSum@f, 0, 6, 3D
The sum = - 32.
1
2
3
4
5
6
-2
-4
-6
-8
The right sum with three subintervals is equal to f H2L × Dx + f H4L × Dx + f H6L × Dx = -32.
The midpoint sum will involve three rectangles where the heights are computed at the midpoints of the subintervals, so it equals
f H1L × D + f H3L × Dx + f H5L × Dx = H-5L * 2 + H-9L * 2 + H-5L * 2 = -38. Here is a picture:
MidpointSum@f, 0, 6, 3D
The sum = - 38.
1
2
3
4
5
6
-2
-4
-6
-8
The three Riemann Sums using six subintervals are computed in the same way; we record the answers here without further
comment: The left sum = -35, the right sum = -35, and the midpoint sum = -36.5.
9
10
final review answers.nb
15. Find the exact value of Ù0 Ix 2 - 6 xM â x using the Fundamental Theorem of Calculus.
6
An antiderivative of the integrand is FHxL =
x3
3
- 3 x2 , so Ù0 Ix2 - 6 xM â x = FHxL
6
6
0
= FH6L - FH0L = -36.
16. Use geometry to find Ù0 f HxL â x, where f is the function whose graph is shown.
5
y = f @xD
2.0
1.5
1.0
0.5
1
2
3
4
5
We add the areas of the following regions, read from left to right: triangle + square + rectangle with semicircular dome on top +
triangle = 1/2 + 1 + (2 + Π/2) + 1/2 = 4 + Π/2. Let's compare this to a Riemann Sum approximation (the "internal" name of this
function is f4).
N@4 + Π  2D
5.5708
MidpointSum@f4, 0, 5, 40D
The sum = 5.57834
2.0
1.5
1.0
0.5
1
2
3
4
5
17. Compute the exact values of the following integrals.
In each case, we use the Fundamental Theorem of Calculus, and compare with a midpoint sum:
a) Ù1
31
x
â x = lnHxL
3
1
= lnH3L - lnH1L =1.09861-0 = 1.09861.
final review answers.nb
f@x_D := 1  x
MidpointSum@f, 1, 3, 40D
The sum = 1.09852
1.0
0.8
0.6
0.4
0.2
1.5
b) Ù-Π cosHxL â x = sinHxL
А2
А2
-Π
2.0
2.5
3.0
= sinHΠ  2L - sinH-ΠL = 1 - 0 = 1.
MidpointSum@Cos, - Π, Π  2, 50D
The sum = 1.00037
1.0
0.5
-3
-2
-1
1
-0.5
-1.0
c) aÙ1
4
x â x = Ù1 x12 â x =
4
f@x_D :=
x
2
3
x32
4
1
=
16
3
-
2
3
=
14
3
= 4.666 ...
11
12
final review answers.nb
MidpointSum@f, 1, 4, 50D
The sum = 4.6667
2.0
1.5
1.0
0.5
1.5
2.0
2.5
3.0
3.5
4.0