Math 220
Exam 2 Sample Problems
Solution Guide
October 25, 2013
√
1. Consider the formula f (x) = 2 − x + 4.
(a) What is the largest subset of A ⊆ R so that f : A → R defined by f (x) = 2 −
a function?
√
x + 4 is
(b) Compute the image of f : A → R.
(c) Compute f ([5, 12)).
(d) Compute f −1 ([0, 2]).
Solution: (a) The formula for f (x) does not take values in R if x + 4 < 0, so if we take
√
A = [−4, ∞), then f (x) = 2 − x + 4 defines a function f : A → R.
√
(b) We know that for any x ≥ 0, 0 ≤ x < ∞, so if x ∈ [−4, ∞, then
√
√
−∞ < − x + 4 ≤ 0 ⇔ −∞ < 2 − x + 4 ≤ 2.
Therefore, the image of f is (−∞, 2].
(c) From the graph of f (x), we know that f is a decreasing function for all x ∈ [−4, ∞. We
know that
√
√
f (5) = 2 − 9 = −1, f (12) = 2 − 16 = −2,
and so by the intermediate value theorem, f ([5, 12)) = (−2, −1].
(d) We use similar reasoning in as in (c). We see easily that
f (0) = 0,
f (−4) = 2.
The inverse image image of [0, 2] will be all x values that are sent to [0, 2] by f . We see then
that f −1 ([0, 2]) = [−4, 0].
2. Let f : R → R be defined by f (x) = x4 + x2 .
(a) Compute the image of f . Explain your answer.
(b) Compute f ([−1, 2]). Explain your answer.
Solution: (a) The image of f is [0, ∞). The reason is that f (x) = x4 + x2 is a polynomial,
whose values are clearly non-negative. Since f (0) = 0, we see that 0 ∈ Im(f ). As the values
of f go to ∞ as x → ±∞, we see from the intermediate value theorem that f takes on all
values between 0 and ∞, so Im(f ) = [0, ∞).
(b) We see that
f (−1) = 1 + 1 = 2,
f (2) = 16 + 4 = 20,
so this might lead us to believe that f ([−1, 2]) = [2, 20]. However, we know that f (0) =
0, and 0 ∈ [−1, 2], so in fact the values of f can be as low as 0 on [−1, 2]. Therefore,
f ([−1, 2]) = [0, 20].
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3. Determine if the given function is injective, surjective, and/or bijective. Which of these
functions are invertible? No formal proof required.
(a) f : R → [0, 1] defined by f (x) = 1 − sin2 (x).
(b) g : [0, π2 ] → [−1, 1] defined by g(x) = cos(x).
(c) h : Z → 3Z defined by h(n) = 3(n + 1).
(d) j : R × R → [0, ∞) defined by j(x, y) = (x + y + 1)2 .
Solution: (a) The function repeats several values; for example f (0) = f (π) = 1, so f is
not injective. Since f (0) = 1 and f (π/2) = 0, the intermediate value theorem implies that
every value between 0 and 1 is in the image of f . So f is surjective.
(b) The values of cos(x) are non-negative for x ∈ [0, π2 ], so g is not surjective. However, g is
decreasing on [0, π2 ], so g is injective.
(c) Every integer multiple of 3 can be expressed as 3(n+1) for some n ∈ Z, so h is surjective.
It is also not hard to show that h is injective, and so h is bijective. Bijective functions are
also invertible, so h is invertible.
(d) We note that j(0, −1) = 0, and similarly we can show that j(x, y) takes on every positive
value, so j is surjective. However, j(0, −1) = j(−1, 0) = 0, so j is not injective.
4. Define f : R − {1} → R − {2} by f (x) =
2x − 1
.
x−1
(a) Show that f is injective.
(b) Show that f is surjective.
(c) Find f −1 (x).
Solution: (a) Suppose that f (x1 ) = f (x2 ) for x1 , x2 ∈ R − {1}. Then
2x1 − 1
2x2 − 1
=
⇔ (2x1 − 1)(x2 − 1) = (2x2 − 1)(x1 − 1)
x1 − 1
x2 − 1
⇔ 2x1 x2 − 2x1 − x2 + 1 = 2x1 x2 − x1 − 2x2 + 1
⇔ −x1 = −x2 ,
and thus x1 = x2 . Therefore f is injective.
(b) Let y ∈ R − {2}. We want to find x ∈ R − {1} so that f (x) = y. That is, we want to solve
2x − 1
=y
x−1
for x. Well,
2x − 1
= y ⇔ 2x − 1 = (x − 1)y
x−1
⇔ 2x − 1 = xy − y
⇔ 2x − xy = 1 − y
⇔ x(2 − y) = 1 − y.
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Thus x = 1−y
2−y . Since y 6= 2, this value of x is well-defined. Since 1 − y 6= 2 − y for any y, the
value f (x) is never 1. One checks that
1−y
= y,
f
2−y
and so f is surjective.
(c) By the argument from part (b), we see that
f −1 (x) =
1−x
.
2−x
5. Let f : A → B and g : B → C be functions. Suppose that f is surjective and that g is not
injective.
(a) Translate “f is surjective” into mathematical symbols.
(b) Translate “g is not injective” into mathematical symbols.
(c) Prove that g ◦ f : A → C is not injective.
Solution: (a) ∀b ∈ B, ∃a ∈ A, (f (a) = b).
(b) ∃b1 , b2 ∈ B, (b1 6= b2 and g(b1 ) = g(b2 )).
(c) Because g is not injective, we can find b1 , b2 ∈ B so that b1 6= b2 and g(b1 ) = g(b2 ).
Because f is surjective, we can find a1 ∈ A so that f (a1 ) = b1 , and also a2 ∈ A so that
f (a2 ) = b2 . Then a1 6= a2 , since f (a1 ) 6= f (a2 ), but
(g ◦ f )(a1 ) = g(f (a1 )) = g(b1 ) = g(b2 ) = g(f (a2 )) = (g ◦ f )(a2 ).
Therefore g ◦ f is not injective.
6. Let f : A → B be a function. Let W ⊆ B.
(a) Prove that f (f −1 (W )) ⊆ W .
(b) Prove that if f is surjective then f (f −1 (W )) = W .
Solution: (a) Let z ∈ f (f −1 (W )). Therefore, z = f (a) for some a ∈ f −1 (W ). But by the
definition of inverse image, a ∈ f −1 (W ) if and only if f (a) ∈ W . Since z = f (a), this implies
that z ∈ W . Therefore f (f −1 (W )) ⊆ W .
(b) Suppose that f is surjective. Using part (a), it suffices to show that W ⊆ f (f −1 (W )).
Let w ∈ W be arbitrary. Since f is surjective, there is some a ∈ A so that f (a) = w. Since
w ∈ W , this implies that f (a) ∈ W , and so a ∈ f −1 (W ). Therefore w = f (a) is in f (f −1 (W )).
Therefore W ⊆ f (f −1 (W )), and so W = f (f −1 (W )).
Problems on topics from Chapter 4: Since the homework due on the day of the exam is from
Chapter 4, studying the assigned problems and others like them from the book would be good
examples.
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