California State University Northridge
MATH 255A: Calculus for the Life Sciences I
Midterm Exam 1
Feb 27, 2013. Duration: 75 Minutes. Instructor: Jing Li
Solutions
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1
Student number:
Problem
1
2
3
4
5
6
7
8
Points
5
10
15
10
15
6
10
12
Your Marks
Question 1. [5 points] Let f (x) = 1 − x 2 and g (x) = ln x. Find
(a)[2 points] f (g (x)) and its domain;
Solution:
¡
¢
f g (x) = 1 − (ln x)2 ,
its domain is (0, ∞).
(b)[3 points] g ( f (x)) and its domain.
Solution:
¡
¢
g f (x) = ln(1 − x 2 ),
from which, we have
1 − x 2 > 0 ⇒ x 2 < 1 ⇒ |x| < 1 ⇒ −1 < x < 1.
2
9
(BONUS)
9
Total
83
Question 2. [10 points] 500 g of iodine 131 is decaying exponentially. After 3 days 386 g of
iodine 131 is left.
(a)[6 points] Write an exponential equation to express the amount of iodine 131 as a function
of time;
Solution: y = y 0 e kt ,
when t = 0, y = 500, so y 0 = 500, i.e., y = 500e kt ;
when t = 3, y = 386.
386 = 500e 3k ⇒
386
ln 0.772
= e 3k ⇒ e 3k = 0.772 ⇒ 3k = ln 0.772 ⇒ k =
500
3
k ≈ −0.0863
Hence,
y = 500e −0.0863t
(b)[4 points] Use your answer from part (a) to find the half-life of iodine 131.
Solution:
ln 21
1
1
1
−0.0863t
−0.0863t
y0 = y0e
⇒ =e
⇒ ln = −0.0863t ⇒ t =
≈ 8.0
2
2
2
−0.0863
The half-life is about 8.0 days.
3
Question 3. [15 points] Use properties of Logarithms:
p
2
x +1
(a)[5 points] Expand log (x+2)
(x+1)2 (x−3)
Solution:
p
p
¢
¡
¢
¡
(x + 2) x 2 + 1
2 + 1 − log (x + 1)2 (x − 3)
x
log
=
log
(x
+
2)
(x + 1)2 (x − 3)
p
¡
¢
= log(x + 2) + log x 2 + 1 − log(x + 1)2 + log(x − 3)
p
= log(x + 2) + log x 2 + 1 − log(x + 1)2 − log(x − 3)
1
= log(x + 2) + log(x 2 + 1) − 2 log(x + 1) − log(x − 3).
2
(b)[5 points] Find the integer value of (log3 81)(log4 64)
Solution:
(log3 81)(log4 64) = log3 34 log4 43 = (4 log3 3)(3 log4 4) = 4 · 3 = 12.
(c)[5 points] Solve for x:
log2 x = 2 − log2 (4 − x)
Solution: log2 x + log2 (4 − x) = 2 so that log2 x(4 − x) = 2 and
4x − x 2 = x(4 − x) = 22 = 4.
So (x − 2)2 = x 2 − 4x + 4 = 0 ⇒ x = 2.
4
Question 4. [10 points] Let f (x) = 2 cos(πx) − 1.
(a)[4 points] Find f (0), f ( 13 ), f ( 12 ), f (1);
Solution:
f (0) = 2 cos 0 − 1 = 2 − 1 = 1
π
1
1
f ( ) = 2 cos( ) − 1 = 2 − 1 = 1 − 1 = 0
3
3
2
1
π
f ( ) = 2 cos( ) − 1 = 0 − 1 = −1
2
2
f (1) = 2 cos(π) − 1 = 2(−1) − 1 = −3.
(b)[4 points] Graph f (x) over an interval of one period;
Solution:
(c)[2 points] Find the maximum and minimum values of the function.
Solution:
max = 2 and min = −2.
5
Question 5. [15 points] Find the limits
x 2 − 3x − 10
x→5
x −5
(a) [3 points] lim
Solution:
x 2 = 3x − 10
(x − 5)(x + 2)
= lim
= lim (x + 2) = 5 + 2 = 7.
x→5
x→5
x→5
x −5
x −5
lim
x2 − 9
x→−3 x 2 + x − 6
(b) [3 points] lim
Solution:
x2 − 9
(x − 3)(x + 3)
(x − 3) −3 − 3 6
=
lim
=
lim
=
= .
x→−3 x 2 + x − 6
x→−3 (x − 2)(x + 3)
x→−3 (x − 2)
−3 − 2 5
p
x −6
(c) [3 points] lim
x→36 x − 36
lim
Solution:
p
p
x −6
x −6
1
1
1
1
= lim p
= .
lim
= lim p
=p
=
p
x→36 x − 36
x→36 ( x − 6)( x + 6)
x→36 x + 6
36 + 6 6 + 6 12
x 2 + 2x − 5
x→∞ 3x 2 + 2
(d) [2 points] lim
Solution:
x 2 + 2x − 5
lim
= lim
x→∞ 3x 2 + 2
x→∞
x2
x2
+ 2x
− x52
x2
3x 2
x2
+ x22
= lim
1 + x2 − x52
x→∞
3+
2
x2
=
1+0−0 1
=
3+0
3
2x 2 − 1
x→∞ 3x 4 + 2
(e) [2 points] lim
Solution:
2x 2 − 1
lim
= lim
x→∞ 3x 4 + 2
x→∞
2x 2
x4
3x 4
x4
− x14
+ x24
= lim
x→∞
2
x2
− x14
3+
2
x4
=
0−0 0
= = 0.
3+0 3
x 4 − x 3 − 3x
x→∞
7x 2 + 9
(f) [2 points] lim
Solution:
x 4 − x 3 − 3x
lim
= lim
x→∞
x→∞
7x 2 + 9
x4
x4
3
− xx 4 − 3x
x4
7x 2
x4
+ x94
= lim
x→∞
1 − x1 − x33
7
x2
+
9
x4
=
1−0−0 1
= .
0+0
0
Division by 0 is undefined, so this limit does not exist. By examining what happens as larger
4
−x 3 −3x
and larger values of x are put into the function, we see that limx→∞ x 7x
= ∞.
2 +9
6
Question 6. [6 points] Let

if x < 0
 0
2
x
−
5x
if
0≤x ≤5
f (x) =

5
if x > 5
(a)[3 points] Graph the above function;
Solution:
(b)[1 point] Find all values of x where the function is discontinuous;
Solution:
f (x) is discontinuous at x = 5.
(c)[2 points] Find the limit from the left and from the right at any values of x found in part
(b).
Solution:
limx→5− f (x) = limx→5− (x 2 − 5x) = 52 − 5 · 5 = 0.
limx→5+ f (x) = limx→5+ 5 = 5.
7
Question 7. [10 points] Epidemiologists estimate that t days after the flu begins to spread in
some college town, the percent of the population infected by the flu is approximated by
p(t ) = t 2 + t
for 0 <= t <= 5.
(a)[4 points] Find the average rate of change of p with respect to t over the interval from 1
to 4 days;
Solution:
The average rate is equal to
p(4) − p(1) (42 + 4) − (12 + 1) 20 − 2 18
=
=
=
= 6.
4−1
4−1
3
3
(b)[6 points] Find the instantaneous rate of change of p with respect to t at t = 3.
Solution: Consider limh→0
We have
p(t +h)−p(t )
h
at t = 3.
p(3 + h) − p(3)
h→0
h
(3 + h)2 + (3 + h) − (32 + 3)
= lim
h→0
h
2
2
3 + 6h + h + 3 + h − 12
= lim
h→0
h
h 2 + 7h
= lim
= lim (h + 7)
h→0
h→0
h
= 7.
lim
8
2
Question 8. [12 points] Let f (x) = 1 − x2 .
(a)[6 points] Use the definition of the derivative to find f 0 (x);
Solution:
(1 −
f (x + h) − f (x)
f (x) = lim
= lim
h→0
h→0
h
0
2
− (x+h)
2
x2
2
2
(x+h)2
) − (1 − x2 )
2
h
2
= lim
2
1 − (x+h)
− 1 + x2
2
h→0
h
x 2 − (x + h)2
x 2 − x 2 − 2xh − h 2
= lim
h→0
h→0
h→0
h
2h
2h
2
−2xh − h
−2x − h
= lim
= lim
= −x.
h→0
h→0
2h
2
= lim
+
= lim
(b)[4 points] Find the equation of the tangent line to the graph at x = −2;
Solution:
2
0
y − y 0 = m(x − x 0 ). x 0 = −2 so that y 0 = f (x 0 ) = 1− (−2)
2 = 1−2 = −1. m = f (x 0 )− x 0 = 2. Also,
y − (−1) = 2(x − (−2)) ⇒ y + 1 = 2(x + 2) ⇒ y = 2x + 3.
(c)[2 points] Sketch the tangent line from part (b) on the graph below. Find the y-intercept.
Solution:
The y-intercept is 3.
9
Question 9. [BONUS: 9 points]
Sketch the graph of the derivative for the function shown.
y
-6
-4
2
-2
-1
-2
-3
You can sketch your answer in the following graph:
Solution:
10
4
6
x
11