Department of Mathematics, University of Wisconsin-Madison
Math 114
Worksheet Section 5.1-5.3
1. For the given functions f and g, find (a) (f ◦ g)(4), (b) (g ◦ f )(2), (c) (f ◦ f )(1), (d) (g ◦ g)(0).
√
(a) f (x) = x + 1; g(x) = 3x
Solution:
√
i. (f ◦ g)(4) = f (g(4)) = f (12) = 13
√
√
ii. (g ◦ f )(2) = g(f (2)) = g( 3) = 3 3
p√
√
iii. (f ◦ f )(1) = f (f (1)) = f ( 2) =
2+1
iv. (g ◦ g)(0) = g(g(0)) = g(0) = 0
3
(b) f (x) = x 2 ; g(x) =
2
x+2
Solution:
1
1
(a) (f ◦ g)(4) = f (g(4)) = f ( ) = √
3
3 3
√
1
2
=√
(b) (g ◦ f )(2) = g(f (2)) = g( 8) = √
2 2+2
2+1
(c) (f ◦ f )(1) = f (f (1)) = f (1) = 1
(d) (g ◦ g)(0) = g(g(0)) = g(1) =
2
3
2. For the given functions f and g, find (a) (f ◦ g)(x), (b) (g ◦ f )(x), (c) (f ◦ f )(x), (d) (g ◦ g)(x) and
state the domain of each composite function.
(a) f (x) = x + 1; g(x) = x2 + 4
Solution:
i. (f ◦ g)(x) = f (g(x)) = f (x2 + 4) = x2 + 5, the domain is R.
ii. (g ◦ f )(x) = g(f (x)) = g(x + 1) = x2 + 2x + 5, the domain is R.
iii. (f ◦ f )(x) = f (f (x)) = f (x + 1) = x + 2, the domain is R.
iv. (g ◦ g)(x) = g(g(x)) = g(x2 + 4) = x4 + 8x2 + 20, the domain is R.
1
(b) f (x) = x2 + 4; g(x) =
√
x−2
Solution:
√
i. (f ◦ g)(x) = f (g(x)) = f ( x − 2) = x + 2, the domain is [2, ∞).
√
ii. (g ◦ f )(x) = g(f (x)) = g(x2 + 4) = x2 + 2, the domain is R.
iii. (f ◦ f )(x) = f (f (x)) = f (x2 + 4) = x4 + 8x2 + 20, the domain is R.
p√
√
x − 2 − 2, the domain is [6, ∞).
iv. (g ◦ g)(x) = g(g(x)) = g( x − 2) =
3. For the following functions, find f and g so that (f ◦ g) = H.
(a) H(x) = (1 + x2 )3
Solution: f (x) = x3 , g(x) = 1 + x2
(b) H(x) = |2x2 + 3|
Solution: f (x) = |x|, g(x) = 2x2 + 3
4. For the given functions f and g, verify that the functions are inverse of each other by showing that
f (g(x)) = x and g(f (x)) = x. Give any values that need to be excluded from the domain of f and the
domain of g.
(a) f (x) = 3 − 2x; g(x) = − 12 (x − 3)
Solution:
1
1
g(f (x)) = g(3 − 2x) = − (3 − 2x − 3) = − (−2x) = x
2
2
,
1
1
f (g(x)) = f (− (x − 3)) = 3 − 2(− (x − 3)) = 3 + (x − 3) = x
2
2
(b) f (x) =
2x + 3
4x − 3
; g(x) =
x+4
2−x
Solution:
4x − 3
2
+3
2(4x − 3) + 3(2 − x)
5x
4x − 3
f (g(x)) = f (
)= 2−x
=
=
=x
4x − 3
2−x
(4x − 3) + 4(2 − x)
5
+4
2−x
2x + 3
4
−3
2x + 3
4(2x + 3) − 3(x + 4)
5x
g(f (x)) = g(
)= x+4
=
=
=x
2x
+
3
x+4
2(x + 4) − (2x + 3)
5
2−
x+4
2
5. In the following f is one to one. (a) Find its inverse function f −1 and check your answer. (b) Find the
domain and the range of f and f −1 .
(a) f (x) = −
3x + 1
x
3x + 1
Solution: Let y = f (x) = −
, then −xy = 3x+1, this is equivalent with −(3+y)x = 1,
x
1
1
therefore x = −
. Hence the inverse for f is f −1 (x) = −
. Domain is (−∞, −3) ∪
3+y
3+x
(−3, ∞), range is (−∞, 0) ∪ (0, ∞)
(b) f (x) =
x2 + 3
;x>0
3x2
x2 + 3
, then 3x2 y = x2 + 3, this is equivalent with x2 (3y − 1) = 3,
3x2
r
3
3
hence x2 =
. Since x > 0, we have x =
. Therefore the inverse is f −1 (x) =
3y
−
1
3y
−
1
r
3
. The domain is ( 13 , ∞), range is (0, ∞).
3x − 1
Solution: Let y = f (x) =
6. For the following, use transformations to graph each function. Determine the domain, range, and
horizontal asymptote.
(a) f (x) = 1 − 2x+3
Solution: First we shift 2x to left by 3, then reflect it with respect to y-axis, finally we move
it up by 1, then we get
50
−4
−2
2
4
−50
The domain is R, range is (−∞, 1), the horizontal asymptote is y = 1.
(b) f (x) = ex+2 − 4
3
Solution: First we shift the graph of ex to left by 2, then shift it down by 4, we get:
50
−4
−2
2
4
−50
The domain is R, range is (−4, ∞), horizontal asymptote is y = −4.
7. Solve each equation.
(a) 8−x+14 = 16x
Solution: Left side is 23(−x+14) , right hand is 24x , hence the equation is equivalent with
3(−x + 14) = 4x, x = 6.
2
(b) (e4 )x · ex = e12
2
2
Solution: Left side is e4x · ex = ex +4x . Hence the equation is equivalent with x2 + 4x = 12,
hence x2 + 4x − 12 = 0, factor this, we have (x − 2)(x + 6) = 0, hence x = 2 or −6.
4