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Math 10850, Honors Calculus 1
Homework 4
Solutions
General and specific notes on the homework
All the notes from homework 1 still apply! Also, please read my emails from September 6
and 13 with comments on the first homework and second homework.
Reading for this homework
Chapters 3 and 4 of Spivak (The homework is on both).
Assignment
1. Spivak Chapter 3 Question 1, parts (i), (iii) [for both of these parts, explicitly say what
is the domain of the new function], (vi) and (vii)
Solution (part i):
f(f(x)) = 1 = 1 =1+x (this last equality valid as long as x 6= −1).
1+f(x) 1+ 1 2+x
1+x
The domain is {x ∈ R : x 6= −1,−2}. bf NB: −1 is not in the domain, even though
the final expression makes perfect sense at x = −1. The domain of the composition
g ◦ h is the set of all domain elements of the first function (h, or in this case f), whose
outputs are in the domain of the second function (g, or in this case f); and −1 is not
in the domain of the first function.
Solution (part iii):
f(cx) = 1 .
1+cx
The domain is R\{−1/c} (or (−∞,−1/c)∪(−1/c,∞) or {x ∈ R : x 6= −1/c}).
Solution (part vi): Given c, we want to find an x such that
1 = 1
1+cx 1+x
with x 6= −1,−1/c. This is equivalent to cx = x or (c − 1)x = 0, which can be solved
for all c by taking x = 0.
1
Solution summary: all c ∈ R.
Solution (part vii): As in the last part, for given c ∈ R we are trying to solve
(c − 1)x = 0 with x 6= −1,−1/c. If c 6= 1 then c − 1 6= 0 so the only solution to
(c −1)x = 0 is x = 0. If c = 1 then the equation becomes 0.x = 0, which has infinitely
many solutions (and so certainly has two different solutions).
Solution summary: only c = 1.
2. Spivak Chapter 3 Question 2, parts (i), (iii), (v)
Solution (part i):
• For all irrational numbers y greater than 1, 1 = h(y) ≤ y. (Good!)
• For all irrational numbers y less than 1, 1 = h(y) > y. (Bad :()
• For all rational numbers y greater than or equal to 0, 0 = h(y) ≤ y. (Good!)
• For all rational numbers y less than 0, 0 = h(y) > y. (Bad :()
So h(y) ≤ y if and only if either: y is an irrational number greater than 1, or: y is a
rational number than or equal to 0.
Solution (part iii): If z is rational then h(z) = 0 so g(h(z)) = 0 and g(h(z))−h(z) = 0.
If z is irrational then h(z) = 1 so g(h(z)) = 1 and g(h(z)) − h(z) = 1 − 1 = 0.
So g(h(z))−h(z) = 0 for all z.
4 4 2
Solution (part v): g(g(ε)) = ǫ , so we seek to find all ε with ǫ = ε . This has three
solutions, ε = 0, ε = 1 and ε = −1.
3. Spivak Chapter 3 Question 3, parts (ii), (iii) and (iv)
√ 2 2
Solution (part ii): In order for the inner 1−x to make sense, we need 1−x ≥ 0,
√ 2 √ 2
or x ∈ [−1,1]. For all such x, we have 0 ≤ 1−x ≤1,so1≥1− 1−x ≥0. This
p √ 2
means that for all x ∈ [−1,1], the expression 1− 1−x makessense, and so the
domain of this last function is [−1,1].
Solution (part iii): The domain is R\{1,2}.
√ 2 2
Solution (part iv): In order for 1−x to make sense, we need 1−x ≥ 0, or x ∈
√ 2 2
[−1,1]. In order for x −1tomakesense,weneedx −1≥0,orx∈(−∞,−1]∪[1,∞).
In order for both to make sense simultaneously, we need x to be in the intersection of
these two sets, that is, we need x = 1 or −1. So the domain is {−1,1}.
4. Spivak Chapter 3 Question 5, parts (ii), (iv) and (vii)
Solution (part ii): s ◦ P
Solution (part iv): S ◦ s
Solution (part vii): s ◦ s ◦ s ◦ P ◦ P ◦ P ◦ s
2
5. Spivak Chapter 3 Question 9 (your answer to part b) should be very short)
Solution (part a):
• CA∩B = CA×CB
• CA∪B = CA+CB−CA×CB
• CR−A = 1−CA
Solution (part b): Take A = {x ∈ R : f(x) = 1}. Then f = CA.
2 2
Solution (part c): Suppose that f = f . Then for each real x, f(x) = (f(x)) , so
either f(x) = 0 or f(x) = 1. By part b), f = CA for some A.
Suppose on the other hand that f = C for some A. Then if x ∈ A, f(x) = 1 = 12 =
A
2 2 2 2 2
f (x), while if x 6∈ A, f(x) = 0 = 0 = f (x). In either case f(x) = (f(x)) , so f = f .
6. Spivak Chapter 3 Question 12
Solution (part a):
• f, g both even: (f + g)(−x) = f(−x) + g(−x) = f(x) + g(x) = (f + g)(x), so
f +g is even in this case.
• f even, g odd: (f +g)(−x) = f(−x)+g(−x) = f(x)−g(x) = (f −g)(x). This
suggests that in general f + g is neither even nor odd in this case. For example,
2 2
take f(x) = x and g(x) = x, so (f +g) = x +x. This is neither even nor odd
(f(−1) = 0, which is equal to neither f(1) nor −f(1)).
• f odd, g even: Same as last case, by symmetry; in general f +g is neither even
nor odd in this case.
• f, g both odd: (f +g)(−x) = f(−x)+g(−x) = −f(x)−g(x) = −(f +g)(x), so
f +g is odd in this case.
Solution (part b):
• f, g both even: (fg)(−x) = f(−x)g(−x) = f(x)g(x) = (fg)(x), so fg is even in
this case.
• f even, g odd: (fg)(−x) = f(−x)g(−x) = f(x)(−g(x)) = −(fg)(x), so fg is odd
in this case.
• f odd, g even: Same as last case, by symmetry; fg is odd in this case.
2
• f, g both odd: (fg)(−x) = f(−x)g(−x) = (−f(x))(−g(x)) = (−1) (fg)(x) =
fg(x), so fg is even in this case.
Solution (part c):
• f, g both even: (f ◦g)(−x) = f(g(−x)) = f(g(x)) = (f ◦g)(x), so f ◦g is even in
this case (notice that we did not use that f is even in this case — if g is even and
f is anything then f ◦ g is even).
3
• f even, g odd: (f ◦ g)(−x) = f(g(−x)) = f(−g(x)) = f(g(x)) = (f ◦ g)(x), so
f ◦ g is even in this case (here we used properties of both f and g).
• f odd, g even: (f ◦ g)(−x) = f(g(−x)) = f(g(x)) = (f ◦g)(x), so f ◦ g is even in
this case (we could have deduced this from our analysis of the case where both
are even).
• f, g both odd: (f ◦g)(−x) = f(g(−x)) = f(−g(x)) = −f(g(x)) = −(f ◦g)(x), so
f ◦ g is odd in this case (here we used properties of both f and g).
Solution (part d): Let c be any real number, and define g : R → R as follows:
c
g (x) = f(x) if x ≥ 0
c c if x < 0.
If x ≥ 0 then g (|x|) = g (x) = f(x), while if x < 0 then g (|x|) = g (−x) = f(−x)
c c c c
(note that −x ≥ 0 in this case), and f(−x) = f(x) (since x is even), so g (|x|) = f(x).
c
So for all real x, g (|x|) = f(x).
c
Clearly there are infinitely many (different) functions of the form f , since there are
c
infinitely many c.
The key point here is: since g(|x|) only uses the rule for g on positive inputs, we can
do whatever we like on negative inputs.
7. Spivak Chapter 3 Question 21
Solution (part a): This is false in general. Consider, for example, g = h both being
the identity function (that sends x to x) and f being the function that sends x to x2.
Wehave
(f ◦ (g + h))(x) = f((g + h)(x)) = f(g(x) + h(x)) = f(x + x) = f(2x) = 4x2
and
2 2 2
((f◦g)+(f◦h))(x) = (f◦g)(x)+(f◦h)(x) = f(g(x))+f(h(x)) = f(x)+g(x) = x +x = 2x ,
and the two (4x2 and 2x2) are not in general equal.
Solution (part b): We have
((g +h)◦f)(x) = (g+h)(f(x))
= g(f(x))+h(f(x))
= (g◦f)(x)+(h◦f)(x)
= ((g◦f)+(h◦f))(x)
so (g + h) ◦ f = (g ◦ f) + (h ◦ f).
4
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