HW 6 Solutions
Math 115, Winter 2009, Prof. Yitzhak Katznelson
√
17.1. f (x) = 4 − x for x ≤ 4 and g(x) = x2 for all x ∈ R.
a) Give the domains of f + g, f g, f ◦ g and g ◦ f .
f + g has domain {x : x ≤ 4}, as does f g, since both of these have
domain dom(f )∩dom(g). The domain of f ◦ g is the set of x with x
in the domain of g and g(x) in the domain of f , so this requires that
x ∈ R and x2 ≤ 4; i.e. that x ∈√[−2, 2]. Similarly, the domain of g ◦ f
is the set of x where x ≤ 4 and 4 − x ∈ R; of course, that square root
is in R as long as it makes sense, i.e. as long as x ≤ 4, so the domain
is {x : x ≤ 4}. (Naively, you might compute that g ◦ f = 4 − x and
thus conclude that the domain is R, but that would be wrong, since
g ◦ f means to first apply f and then apply g, and we cannot apply f
to anything greater than 4).
b) f ◦ g(0) = f (0) √
= 2. g ◦ f (0) = g(2)
√ = 4.
f ◦ g(1) = f (1) = 3. g ◦ f (1) = g(
√ 3) = 3.
f ◦ g(2) = f (4) = 0. g ◦ f (2) = g( 2) = 2.
c) No, f ◦ g 6= g ◦ f ; just look at the values we just computed!
d) The domain of f ◦ g is [−2, 2], so f ◦ g(3) does not make sense.
The domain of g ◦ f is (−∞, 4], so g ◦ f (3) does make sense and in fact
equals g(1) = 1.
17.2. f (x) = 4 for x ≥ 0, f (x) = 0 for x < 0, and g(x) = x2 for all
x. Thus dom(f ) =dom(g) = R.
a) Determine the functions f + g, f g, f ◦ g and g ◦ f . Be sure to
specify their domains.
Well, since f and g both have domain all of R, all four of the functions
listed have domain R as well. (A quotient of the two functions might
pose problems, but we’re not considering that). And computing, we
see that:
(f + g)(x) = x2 + 4 for x ≥ 0, and (f + g)(x) = x2 for x < 0.
(f g)(x) = 0 for x ≤ 0 and (f g)(x) = 4x2 for x ≥ 0.
f ◦ g(x) = 4 for x2 ≥ 0 and 0 for x2 < 0; but x2 ≥ 0 always, so
f ◦ g(x) = 4 on all of R.
g ◦ f (x) = 42 = 16 when x ≥ 0, and g ◦ f (x) = 02 = 0 when x ≤ 0.
b) Which of the above functions is continuous?
We can see that all of the functions are continuous away from x = 0,
because they are polynomials. At x = 0, the limit of the function
needs to exist and equal the value of the function. So: f + g is not
continuous at 0, because the limit of a sequence from the left will be 4
1
2
and the imit of a sequence from the right will be 0, and thus the limit
limx→0 (f + g)(x) does not exist.
However, f g is continuous at 0 - the limits of any sequences approaching 0 from the left and right will both be 0, which agrees with
f g(0) = 0. (This isn’t quite a proof). So f g is continuous on all of R.
f ◦ g is just the constant function 4, which is continuous on all of R.
And g ◦ f is not continuous at 0, because the limit from the left is 0
and the limit from the right is 16. 17.3. Accept on faith that sin x, cos x, exp x, 2x , loge x for x > 0,
xp for p ∈ R and x > 0 are all continuous, then prove the following
functions are continuous:
e) tan x for x 6= odd multiple of π/2. Well, tan x = sin x/ cos x.
By Theorem 17.4 ii), since sin x and cos x are both continuous, we
have that their quotient tan x is continuous whenever cos x 6= 0; i.e.
whenever x 6= an odd multiple of π/2. f) x sin(1/x) for x 6= 0. First we claim that 1/x is continuous for all
x 6= 0. Why is that? Well, it’s continuous for all x > 0 by the above
fact with p = −1. And for x < 0, take a sequence xn → x. Then we
also know that −xn → −x. Since −x > 0 and 1/x is continuous for
all x > 0, we have that 1/(−xn ) → 1/(−x). Thus, multiplying by -1,
1/xn → 1/x. This shows that 1/x is continuous for x < 0.
Now 1/x is continuous for x 6= 0 and sin x is continuous for all x.
Thus by Theorem 17.5, sin(1/x) is continuous for x 6= 0. And then,
since x is continuous for all x, Theorem 17.4 ii) tells us that x sin(1/x)
is continuous for x 6= 0. √
17.4. Prove that the function x is continuous on its domain [0, ∞).
Take a point x0 ∈ [0, ∞) and a sequence xn ∈ [0, ∞) converging to
√
√
x0 . By Example 5 in Section 8, we’re told that lim xn = x0 , which
is exactly what we need to show to verify continuity! 17.8. Let f and g be real-valued functions.
a) Show that min(f, g) = 12 (f +g)− 21 |f −g|. To do this, consider two
cases, f ≥ g and f < g. If f ≥ g, then min(f, g) = g. And |f − g| =
f − g, so 21 (f + g) − 12 |f − g| = f /2 + g/2 − f /2 + g/2 = g = min(f, g).
Conversely, if f < g, then min(f, g) = f , and |f −g| = −(f −g) = g−f ,
so 12 (f + g) − 12 |f − g| = f /2 + g/2 − g/2 + f /2 = f = min(f, g). In
either case the identity holds. b) Show that min(f, g) = − max(−f, −g). Again break things into
two cases. If f ≥ g, then −f ≤ −g, so max(−f, −g) = −g, so
3
− max(−f, −g) = −(−g) = g = min(f, g). Conversely, if f < g,
then −f > −g, so − max(−f, −g) = −(−f ) = f = min(f, g). c) Use either a) or b) to prove that if f and g are continuous at x0 ,
then so is min(f, g).
I’ll use a). By Theorem 17.4 i), f + g and f − g are continuous at x0 .
By Theorem 17.3, so is |f − g|. Also by Theorem 17.3, so are 12 (f + g)
and 12 |f − g|. By Theorem 17.4 i) again, min(f, g) = 12 (f + g) − 21 |f − g|
is thus continuous at x0 . 17.10. Prove that the following functions are discontinuous at the
indicated points.
a) f (x) = 1 for x > 0 and f (x) = 0 for x ≤ 0, x0 = 0. Take
the sequence xn = 1/n; this converges to x0 . f (xn ) = 1 for all n, so
lim f (xn ) = 1 6= 0 = f (x0 ). Thus f (x) is not continuous at 0. b) g(x) = sin(1/x) for x 6= 0 and g(0) = 0, x0 = 0. Take the
1
sequence xn = 2πn+
π . This sequence converges to 0, as it is bounded
2
in absolute value by 1/n. And g(xn ) = sin(2πn + π2 ) = sin π/2 = 1, so
lim g(xn ) = lim 1 = 1 6= 0 = g(0). Thus g(x) is not continuous at 0. c) sgn(x) = −1 for x < 0, sgn(x) = 1 for x > 0, and sgn(0) = 0,
x0 = 0. Take the sequence xn = 1/n; it converges to 0. As with a),
sgn(xn ) = 1 for all n, so lim sgn(xn ) = 1 6= sgn(0) = 0. Thus sgn(x)
is not continuous at 0. d) P (x) = 15 for 0 ≤ x < 1 and P (x) = 15 + 13n for n ≤ x < n + 1,
x0 a positive integer. Let x0 be any positive integer; P (x0 ) = 15+13x0.
Then consider the sequence xm = x0 −1/m. This sequence converges to
x0 . However, x0 − 1 ≤ xm < x0 for all m, so P (xm ) = 15 + 13(x0 − 1) =
15 + 13x0 − 13 = 2 + 13x0 . Thus lim P (xm ) = 2 + 13x0 6= 15 + 13x0 =
P (x0 ), and so P (x) is not continuous at x0 . 17.12. a) Let f be a continuous real-valued function with domain
(a, b). Show that if f (r) = 0 for each rational number r in (a, b), then
f (x) = 0 for all x ∈ (a, b).
Pick any number x ∈ (a, b). If x is rational, f (x) = 0 by definition. If not, by the density of the rationals, there exists a sequence
of rational numbers rn in (a, b) such that rn converges to x. And notice that f (rn ) = 0, so lim f (rn ) = 0. But since f is continuous,
f (x) = lim f (rn ) = 0. b) Let f and g be continuous real-valued functions on (a, b) such that
f (r) = g(r) for each rational number r in (a, b). Prove that f (x) = g(x)
for all x ∈ (a, b).
4
This is easy; by Theorem 17.4 ii), f − g is a continuous real-valued
function, and we know that (f − g)(r) = 0 for any rational number
r ∈ (a, b). Applying a) to f − g, we see that (f − g)(x) = 0 for any
x ∈ (a, b), and thus f (x) = g(x) for all x ∈ (a, b). 17.13. a) Let f (x) = 1 for rational numbers x and f (x) = 0 for
irrational numbers. Show that f is discontinuous at every x ∈ R.
There are two cases: when
x ∈ Q and when x ∈
/ Q. If x ∈ Q, consider
√
the sequence xn = x + n2 . xn converges to x, and xn is irrational for
√
each n, as if xn ∈ Q, then xn − x ∈ Q, then n(xn − x) = 2 ∈ Q,
contradiction. So f (xn ) = 0 for all n, so lim f (xn ) = 0 6= 1 = f (x).
Thus f (x) is not continuous at any x ∈ Q.
Conversely, if x ∈
/ Q, by the density of the rationals there exists a
sequence of rational numbers rn converging to x. f (rn ) = 1 for all n,
so lim f (rn ) = 1 6= 0 = f (x). Thus f (x) is not continuous at any x ∈
/Q
either, and so f (x) is not continuous at any x ∈ R.
b) Let h(x) = x for rational numbers x and h(x) = 0 for irrational
numbers. Show that h is continuous at x = 0 and at no other point.
First we’ll show h is continuous at x = 0. Pick ǫ > 0. Then we
notice that since h(x) is either equal to x or 0, |h(x)| ≤ |x|. So: let
δ = ǫ. If |x − 0| = |x| < δ, then |h(x) − h(0)| = |h(x)| ≤ |x| < δ = ǫ.
This shows that h is continuous at 0.
Now pick x 6= 0; we’ll show that h is not continuous at x. There
are
√
2
two cases. If x ∈ Q: then, again, take the sequence xn = x + n . As
before, xn is irrational for all n, and xn converges to x. But h(xn ) = 0
for all n, so lim h(xn ) = 0 6= x = h(x). So h is not continuous at x.
Conversely, if x ∈
/ Q, then there exists a sequence of rational numbers
rn converging to x. h(rn ) = rn , so lim h(rn ) = lim rn = x 6= 0 = h(x).
Thus h is not continuous at x. So for any x 6= 0, h is not continuous
at x. 17.14. For each rational number x, write x = pq , where p and q are
integers with no common factors and q > 0, and then define f (x) = 1q .
Also define f (x) = 0 if x is irrational. Show that f is continuous at
every point of R \ Q and discontinuous at each point of Q.
First we’ll show that f is discontinuous
at each point of Q. Let
√
x ∈ Q, and, as before, let xn = x + n2 . Then xn converge to x, xn is
irrational for each n, so f (xn ) = 0 for each n, so lim f (xn ) = 0 6= f (x)
(x is rational so f (x) 6= 0). So f is discontinuous at x.
Now pick x0 ∈
/ Q (notice f (x0 ) = 0), and pick ǫ > 0. For each
n ∈ N, let δn be the distance between x0 and the nearest multiple of
5
1
.
n
Since x ∈
/ Q, δn > 0 for each n. Since ǫ > 0, there exists N ∈ N
such that 0 < N1 < ǫ. Let δ = min(δ1 , δ2 , . . . , δN ); then δ > 0. Suppose
|x−x0 | < δ. Then in particular, x is not a multiple of n1 for any n ≤ N.
So either x is irrational, in which case f (x) = 0, or x is rational but
not a multiple of n1 for any n ≤ N. In that case, f (x) = pq for some
q > N where p and q have no common factors, and thus 0 < f (x) < N1 .
In either case, |x − x0 | < δ implies |f (x) − f (x0 )| = |f (x)| < N1 < ǫ.
This shows that f (x) is continuous at x0 .