Solutions Homework 1
This solution sheet is provided as a sample of what I consider good homework solutions (or –
if these were midterm problems – good exam solutions). You might prefer slightly more or less
detail which both would be fine, but there should not be just a solution, nor do easy algebraic
transformations be decomposed in the most minute detail.
Note that the solutions do not just consist of formulas, but – unless steps are straightforward –
there is a bit of accompanying explanatory text!
(Obviously your homework does not need to be typed up.)
§7.1, Problem 22 We have that y = 12 x − 27 . Solving for x yields x = 2y + 7. Now we swap
the role of x and y and get f −1 (x) = 2x + 7. Domain and range of f −1 are the real numbers. We
verify:
1
7
7
7
f (f −1 (x)) = (2x + 7) − = (x + ) − = x
2
2
2
2
and
7
1
x−
+ 7 = x − 7 + 7 = x.
f −1 (f (x)) = 2
2
2
§7.2, Problem 54
We substitute u = sec x + tan x. Then
du = (sec x tan x + sec2 x)dx = (sec x)(tan x + sec x)dx
and thus sec xdx = u1 du. Thus
sec x
dx =
ln(sec x + tan x)
Z
p
Z
1
√
du
u ln u
Substituting v = ln u yields dv = u1 du and therefore
Z
Z
q
√
1
√
du = v −1/2 dv = 2v 1/2 + C = 2 ln u + C = 2 ln(sec x + tan x) + C.
u ln u
§7.2, Problem 64
Let y =
θ sin θ
√
.
sec θ
Then ln y = ln θ + ln(sin θ) − 12 ln(sec θ) and
d
1 cos θ 1
ln y = +
− tan θ
dθ
θ
sin θ
2
and therefore (logarithmic differentiation)
dy
d
θ sin θ
= y ln y = √
dθ
dθ
sec θ
1 cos θ 1
+
− tan θ .
θ
sin θ
2
§7.3, Problem 44
Z
ln 3
− ln 2
0
e−x dx = −e−x − ln 2 = −e0 + eln 2 = −1 + 2 = 1.
§7.3, Problem 68 As ex is increasing, the function f (x) = 2esin(x/2) has a maximum whenever
sin(x/2) has a maximum. This occurs at x = π + 2k(2π) where k is any integer, In this case
sin(x/2) = 1 and f (x) has the value 2e ∼ 5.43656. Similarly minima occur at x = 3π + 2k(2π) with
sin(x/2) = −1 and f (x) = 2/e ∼ 0.73576.