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MA  - Calculus I for the Life Sciences
Final Exam Solutions
Spring 
-May-
. Consider the funion defined implicitly near (2, 1) by x2 y 2 + 3xy = 10y.
(a) [ points] Use implicit differentiation to find the derivative at (2, 1).
i d
d h 2 2
x y + 3xy =
[10y]
dx
dx
dy
dy
dy
+ 3y + 3x
= 10
2xy 2 + 2yx2
dx
dx
dx
dy
dy
dy
2yx2
+ 3x
− 10
= −2xy 2 − 3y
dx
dx
dx
2xy 2 + 3y
dy
=−
dx
2yx2 + 3x − 10
Substituting x = 2 and y = 1, we get
dy 4+3
=−
dx (2,1)
8 + 6 − 10
dy 7
=−
dx
4
(2,1)
(b) [ points] Determine the equation of the tangent line at (2, 1).
7
The equation of the tangent line is y = − (x − 2) + 1 .
4
. Let f (x) be the funion defined by f (x) = e
√
x
for x ≥ 0.
(a) [ points] Write an equation for the line tangent to the graph of f (x) at the
point where x = 1.
√ d √
d √x
1 √
e =e x
x = √ e x,
dx
dx
2 x
e
So the slope of the tangent line at x = 1 is f ′ (1) = . The equation for the
2
e
tangent line is y = (x − 1) + e .
2
f ′ (x) =
-May-
Final Exam
Solutions
(b) [ points] Find the point where this tangent line crosses the x-axis.
The line crosses the x-axis when y = 0.
e
(x − 1) + e = 0
2
e
(x − 1) = −e
2
x − 1 = −2
x = −1
The tangent line crosses the x-axis at (−1, 0).
. Let f be the funion defined by f (x) = x3 − x2 − 2x for 0 ≤ x ≤ 3.
(a) [ points] Find where the graph of f crosses the x-axis. Show your work.
Set f (x) = 0 and solve
x3 − x2 − 2x = 0
x(x2 − x − 2) = 0
x(x + 1)(x − 2) = 0
The solutions are x = −1, x = 0, x = 2, but we are restried to [0, 3], so the
only two places where f crosses the x-axis will be x = 0 and x = 2.
(b) [ points] Find the intervals on which f is increasing.
Find the derivative and the critical points.
f ′ (x) = 3x2 − 2x − 2
Now solve
3x2 − 2x − 2 = 0
√
√
2 ± 28 1 ± 7
x=
=
6
3
√
1+ 7
lies in our interval. We know that f (x) is decreasing if
Only x =
3
√
7
1
+
f ′ (x) < 0 which happens when 0 < x <
. Also, f (x) is increasing if
3
√
1+ 7
f ′ (x) > 0 which happens when
< x < 3.
3
(c) [ points] Find the absolute maximum and absolute minimum value of f .
Justify your answer.
f (x) is continuous on the closed interval [0, 3], so it attains its√maximum and
1+ 7
minimum. There are three points to check: x = 0, x =
, and x = 3.
3

Spring 
MATH  -
Final Exam
-May-
Solutions
From the First
√ Derivative Test we can see that the absolute minimum occurs
1+ 7
at x =
. Evaluating f at  and , we have that f (0) = 0 and f (3) = 12,
3
so the absolute maximum occurs at x = 3.
√
20 + 14 7
The maximum value of f on [0, 3] is  and the minimum value is −
≈
27
−2.1126.
. [ points] A projeile is fired toward the sea from a cliff  feet above the water.
If the launch angle is 45◦ and the muzzle velocity is k, then the height h of the
projeile above the water when it is x feet downrange from the launch site is
h(x) =
−32 2
x + x + 200 feet.
k2
See the figure.
Determine the maximum values of h and the distance downrange where this maximum occurs if k = 2800 feet per second.
h
cliff
x
sea level
32 2
x + x + 200
k2
64
h′ (x) = − 2 x + 1
k
h(x) = −
Set equal to zero and solve
−
64
x+1 = 0
k2
k2
x=
64
Since h′′ (x) < 0, this will be the absolute maximum. Thus, the downrange distance
k2
is x =
= 122, 500 feet. The height at that distance is h(122500) = 61, 450 feet
64

Spring 
MATH  -
Final Exam
-May-
Solutions
. Find the requested quantities in each of the following.
(a) [ points] Let


2 + cos(x − 1) if 0 ≤ x ≤ 1,




f (x) = 
ax − 2
if 1 < x < 2,

 2

x − 2x + b
if 2 ≤ x ≤ 3.
where a and b are constants. Find the values of a and b which make f (x) is
continuous on [0, 3].
We need lim− f (x) = lim+ f (x). This gives the equation
x→1
x→1
2 + cos(0) = a − 2
a=5
Likewise, we need lim− f (x) = lim+ f (x). This gives
x→2
x→2
2a − 2 = 22 − 2(2) + b
8 = 4−4+b
b=8
These values for a and b will make f (x) continuous.
6n2 − 3n + 7
.
n→∞
(n + 3)2
(b) [ points] Evaluate lim
6n2 − 3n + 7
6n2 − 3n + 7
=
lim
n→∞
n→∞ n2 + 6n + 9
(n + 3)2
n2
n
7
6 2 −3 2 + 2
n
n
= lim n2
n→∞ n
n
9
+6 2 + 2
n2
n
n
6
= =6
1
lim
. Find the following limits
sin(x)
.
x→0 x cos(x)
Use l’Hospital’s Rule since both the numerator and denominator go to .
(a) [ points] lim
cos(x)
sin(x)
= lim
x→0 cos(x) − x sin(x)
x→0 x cos(x)
1
=
=1
1−0
lim

Spring 
MATH  -
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Solutions
This can also be done as follows:
sin(x)
sin(x) 1
= lim
x→0 x cos(x)
x→0
x cos(x)
= 1·1= 1
lim
ln(xe x )
.
x→∞
x
Use l’Hospital’s Rule since both the numerator and denominator go to infinity.
(b) [ points] lim
1 x
(e + xe x )
x
ln(xe x )
xe
= lim
lim
x→∞
x→∞
x
1
e x + xe x
= lim
x→∞
xe x
1
= lim + 1 = 1
x→∞ x
x2 − 4
(c) [ points] lim 2
.
x→2 x − x + 6
l’Hospital’s Rule does not apply.
0
x2 − 4
=
x→2 x 2 − x + 6
8
=0
lim
. Find the following general antiderivatives.
Z x2 + x + x−1 + x−2 dx
(a) [ points]
Z 1
1
1
x2 + x + x−1 + x−2 dx = x3 + x2 + ln |x| − + C.
3
2
x
Z 2 sin(x) − 2 cos(x) + 2 sec2 (x) dx
(b) [ points]
Z 2 sin(x) − 2 cos(x) + 2 sec2 (x) dx = −2 cos(x) − 2 sin(x) + 2 tan(x) + C.
(c) [ points]

Z
!
e x − e −x
dx.
2
Z x −x !
Z
e −e
1
1
dx =
(e x − e −x )dx = (e x + e −x ) + C
2
2
2
Spring 
MATH  -
Final Exam
-May-
Solutions
. Use the Fundamental Theorem of Calculus (part ) to evaluate the following derivatives.
Zx
2
cos2 (u)e −u du then F ′ (x) equals
(a) [ points] If F(x) =
1
By the Fundamental Theorem of Calculus, Part I, we know that
Zx
d
2
2
′
cos2 (u)e −u du = cos2 (x)e −x .
F (x) =
dx 1
(b) [ points] If G(x) =
Z
3x2
cos(2u) sin(2u) du then G ′ (x) equals
1
By the Fundamental Theorem of Calculus, Part I, and the Chain Rule we
have
Z 3x2
d
′
G (x) =
cos(2u) sin(2u) du
dx 1
d
(3x2 )
= cos 2(3x2 ) sin 2(3x2 )
dx
= 6x cos(6x2 ) sin(6x2 )
. Find the following definite integrals.
Z 3
1
4x − dx
(a) [ points]
x
1
Z
3
1
4x −
3
1
dx = 2x2 − ln(x)1
x
= (18 − ln 3) − (2 − ln1)
= 16 − ln 3 ≈ 14.9014
(They may use the calculator to compute this integral, but in that case the
answer is either right or wrong — no partial credit.)

Spring 
MATH  -
Final Exam
-May-
Solutions
(b) [ points] The funion h(x) is graphed below.
6
5
4
3
2
1
1
Compute
Z
2
3
4
5
6
7
8
9
10
10
h(x) dx.
0
This is to be done geometrically.
Z
10
h(x) dx =
0
Z
1
h(x) dx +
0
Z
2
h(x) dx +
1
Z
4
h(x) dx +
2
Z
5
h(x) dx +
4
Z
10
h(x) dx
5
1
1
1
= (0)(1) + (2 + 4)(1) + (4 + 2)(2) + (2)(1) + (1 + 5)(5)
2
2
2
= 26
. Find the requested areas.
(a) [ points] Set up (but do NOT evaluate) the integral that will give the area
3π
enclosed by the curves y = cos(x), y = 1+sin(2x) and the vertical line x =
.
2
3
2
1
Π
2
Area =
Z
3π
2
0
Π
3Π
2
[(1 + sin(x)) − cos(x)] dx
3π
≈ 6.71239. They do not have to evaluate the
2
integral, but IF they do, they must do it correly.)
(The corre answer is 2 +

Spring 
MATH  -
Final Exam
-May-
Solutions
√
(b) [ points] The region R is enclosed by y = x/4 and y = x in the first quadrant. Find the area of R.
4
3
2
1
4
8
12
16
√
First, find the points of interseion by setting x = 4x . These are x = 0 and
x = 16. The area is then
Z 16 √
32
x
≈ 10.6667.
x − dx =
AreaR =
4
3
0
(They may use the calculator to compute this integral, but they must show
the integral that they are evaluating and not just the answer.)
BONUS [ points each] Compute the derivatives of the following funions. You do not need
to simplify.
(a) f (x) =
x2 + 1
.
e 2x
f ′ (x) =
2xe 2x − 2(x2 + 1)e 2x
e 4x
(b) f (x) = x5 · sin(2x).
f ′ (x) = 5x4 sin(2x) + 2x5 cos(2x).
(c) f (x) = cos(3x + 5x3 ).
f ′ (x) = −(3 + 15x2 ) sin(3x + 5x3 ).
(d) f (x) = e 2 .
f ′ (x) = 0
(e) f (x) = (4x2 − 1)3 + (8x + 9)11 .
f ′ (x) = 3(4x2 − 1)2 (8x) + 11(8x + 9)10 (8).

Spring 
MATH  -