Download Exam 2 – Answers on Second Page Math 2423 Instructions Give

Exam 2 – Answers on Second Page
Math 2423
Instructions Give thorough explanations of your work to guarantee getting full credit. Use of calculators is not
permitted. Please write on the back of the pages if you need more space.
1. (5 points) Sketch the graphs of y = ex and y = ln(x) in the same picture.
2. (15 points) Determine the intervals of monotonicity for the function f (x) = ln(x2 − 2x + 2) and identify any local
maximums or minimums for f . (Hint: The domain of f consisits of all real numbers.)
3. (5 points) One of the two integrals
Z
ex + sin(x)
dx
ex + cos(x)
Z
and
ex + cos(x)
dx
ex + sin(x)
is easy to compute and the other is very difficult. Identify the easier one and work it out.
(a) sin(tan−1 ( 32 ))
4. (10 points) Use a right triangle analysis to simplify each of:
5. (15 points) If f (x) =
3−x
then find a formula for the inverse function f −1 (x).
5 + 2x
6. (15 points) Differentiate (and simplify):
(a) g(x) = x arctan(x4 ),
7. (20 points) Compute each of the following integrals:
Z
(b)
Z
√
(b) cos(tan−1 ( 23 ))
(a)
R
(b) h(x) = ln(sec(x) + tan(x))
e3x+1 dx
1
dx
1 − x2
e2
1
dx
1/e x
Z p
ln(5x)
(d)
dx
x
(c)
8. (10 points) Let R be the triangular region underneath the line y = 3 − 2x in the first quadrant. If R is rotated
about the x-axis express the volume of the resulting solid as an integral using:
(a) the disk/washer method
(b) the shell method.
9. (5 points) Let R be the triangular region described in the previous problem. If R is rotated about the horizontal
line y = 5 then express the volume of the resulting solid as an integral.
ANSWERS:
1. Since the two functions are inverses of each other, their graphs will be mirror images across the line y = x. Check
the textbook for the graphs of these very important functions.
2. To find the critical points of f we set f 0 (x) = 2(x − 1)/(x2 − 2x + 2) equal to 0, and solve for x. This gives x = 1
as the only critical point. (NOTE: the only way a quotient can equal zero is if its numerator equals zero.) The
intervals of monotonicity are: f is increasing on (1, ∞) and f is decreasing on (−∞, 1). (These are determined by
observing that f 0 (x) > 0 when x > 1, and f 0 (x) < 0 when x < 1 because the numerator x2 − 2x + 2 = (x − 1)2 + 1
is always positive.) There is one local extreme, it is a local minimum at x = 1.
3. The second integral can be worked by making the substitution u = ex + sin(x) (and du = (ex + cos(x))dx), which
results in
Z
ex + cos(x)
dx = ln(| ex + sin(x) |) + C
ex + sin(x)
4. The right traingle would legs with lengths 3 and 2, and the hypotenuse would have length
√
√
2/ 13 and cos(tan−1 ( 23 )) = 3/ 13
5. f −1 (x) =
√
13, so sin(tan−1 ( 32 )) =
3 − 5x
2x + 1
6.
(a) g 0 (x) = arctan(x4 ) +
(b) h0 (x) = sec(x)
4x4
1 + x8
7.
Z
(a)
Z
e3x+1 dx =
1 3x+1
e
+C
3
1
dx = sin−1 (x)
1 − x2
Z e2
1
dx = 3
(c)
x
1/e
Z p
ln(5x)
2
(d)
dx = (ln(5x))3/2 + C
x
3
(b)
√
8. The volume works out to be 9π/2 using either method below (however the problem doesn’t require that the integral
be computed).
(a) Using the disk method, the volume is
(b) Using the shell method, the volume is
R 3/2
R03
0
π(3 − 2x)2 dx.
πy(3 − y) dy.
R 3/2
9. Using the washer method, the volume equals 0 π(25 − (2 + 2x)2 )) dx.
Using the shell method, the volume
R3
equals 0 π(5 − y)(3 − y) dy. (Either integral would be considered to be a correct answer.) With either method
the volume works out to be 18π (however the problem doesn’t require that the integral be computed).