MATH 31B BRIDGE PROGRAM: HW 2 SOLUTIONS
JOE HUGHES
4. Show that f (x) =
range of f and f −1 .
Solution: Set y =
x−2
x+3 ,
x−2
x+3
is invertible, and find its inverse. Determine the domain and
then
(x + 3)y = x − 2
or
xy − x = −2 − 3y
Therefore
x=
3y + 2
1−y
so the inverse is
f −1 (x) =
3x + 2
1−x
f is defined everywhere besides x = −3, so the domain of f (and therefore the range of
f −1 ) is {x 6= −3}.
f −1 is defined everywhere except x = 1, so the domain of f −1 (and therefore the range of
f ) is {x 6= 1}.
12. Find a domain on which f (s) =
restricted to this domain.
1
s2
is invertible, and a formula for f −1 when f is
Solution: f (−s) = f (s), so f is not invertible on the whole real line. But f 0 (s) = −2s−3 < 0
for s > 0, so f is invertible on the interval (0, ∞).
To solve for the inverse, set t =
1
,
s2
so s2 =
1
t
or
1
s = ±√
t
Since s is positive, we should take the positive square root. Therefore
1
f −1 (s) = √
s
20. Show that f (x) =
domain of f .
1
x2 +1
in one-to-one on (−∞, 0] and find a formula for f −1 for this
1
2
JOE HUGHES
Solution: The derivative of f is
f 0 (x) = −(x2 + 1)−2 · 2x =
−2x
(1 + x2 )2
It follows from this formula that f (x) is increasing on the interval (−∞, 0), has a maximum
at x = 0, and is decreasing on (0, ∞). Therefore f is one-to-one on (−∞, 0] (adding in the
extra point 0 doesn’t change anything).
To find the inverse, set y =
1
,
1+x2
so that
x2 + 1 =
1
y
so
p
x = ± y −1 − 1
We should take the negative square root because f was restricted to the domain (−∞, 0].
Note also that the range of f is (0, 1], so y −1 is in the interval [1, ∞), so the term inside
the square root is positive.
Therefore
p
f −1 (x) = − x−1 − 1
31. Let g be the inverse of f (x) = x3 + 2x + 4. Calculate g(7), and g 0 (7).
Solution: First of all, f is increasing because f 0 (x) = 3x2 + 2 > 0, so f is invertible.
Simply by guessing we see that f (1) = 1 + 2 + 4 = 7, hence g(7) = 1. Therefore
1
1
1
1
g 0 (7) = 0
= 0
=
=
f (g(7))
f (1)
3+2
5
32. Find g 0 (− 12 ), where g is the inverse of f (x) =
x3
.
x2 +1
Solution: First of all, the derivative of f is
f 0 (x) =
3x4 + 3x2 − 2x4
x4 + 3x2
3x2 (x2 + 1) − x3 (2x)
=
= 2
2
2
2
2
(x + 1)
(x + 1)
(x + 1)2
which is always non-negative, and is zero only when x = 0. Therefore f is increasing, so is
invertible.
To compute g 0 (− 12 ), we first need to determine g(− 21 ). For this, note that
f (−1) =
−1
1
=−
1+1
2
so g(− 21 ) = −1. Therefore
1
1
(1 + 1)2
4
g0 −
= 0
= 4
= =1
2
2
f (1)
1 +3·1
4
MATH 31B BRIDGE PROGRAM: HW 2 SOLUTIONS
35. If g(x) is the inverse of f (x) =
√
3
x2 + 6x for x ≥ 0, determine g(4) and g 0 (4).
Solution: First observe that x2 + 6x ≥ 0 if x ≥ 0, so f is actually defined on this domain.
Moreover,
2x + 6
x+3
f 0 (x) = √
=√
>0
2 x2 + 6x
x2 + 6x
so f is invertible.
To determine g(4), we need to find the value of x such that f (x) = 4. This is the same as
finding a positive value of x for which x2 + 6x = 16, and either by guessing or using the
quadratic formula we see that x = 2 works. Therefore g(4) = 2.
Finally,
√
1
1
4
22 + 6 · 2
g (4) = 0
= 0
=
=
f (g(4))
f (2)
2+3
5
√
36. If g(x) is the inverse of f (x) = x2 + 6x for x ≤ −6, determine g(4) and g 0 (4).
0
Solution: If x ≤ −6, then x2 ≥ −6x (x is negative so multiplying by x switches the order
of the inequality) or x2 + 6x ≥ 0. So f is defined on this domain. The formula
2x + 6
x+3
f 0 (x) = √
=√
2 x2 + 6x
x2 + 6x
from the previous problem is still valid, except this time f 0 (x) < 0, so f is decreasing.
√
To determine g(4), we need to find a real number x ≤ −6 such that x2 + 6x = 4,
or x2 + 6x = 16. After guessing or using the quadratic formula, we get x = −8. So
g(4) = −8.
Finally,
1
g (4) = 0
=
f (−8)
0
√
82 − 6 · 8
4
=−
3−8
5