Test 1 Practice Problems
√
1) Find the domain of f (x) = 2 − 4x.
1
1
Answer: x ≤ or x in the interval [ , ∞).
2
2
2) Find the x and y intercepts of f (x) = −9 + 3(x + 1)2 .
√
Answer: (−1 ± 3, 0) , (0, −6) are the x-intercepts and y-intercepts,
respectively.
3) Determine whether the function f (x) =
even or odd.
x
is even, odd, or neither
x2 + 3x
Answer: neither
√
4) Find an equation of the final graph of f (x) = x after it is shifted left 1
unit, vertically compressed by a factor of 4 and shifted up 2 units.
1√
Answer: y =
x+1+2
4
5) Find an equation of the line that goes through the points (−3, 4) and
(2, 5).
1
23
Answer: y = x + .
5
5
6) Find an equation of the line through (1, −2) that is perpendicular to the
line x − 2y = −5
Answer: y = −2x

 3x2 + x, x > 2
2,
0 < x ≤ 2 find f (1).
7) If f (x) =
 √
−x + 2, x ≤ 0
Answer: f (1) = 2
2x
2
and g(x) = , find (f ◦ g)(x).
+1
x
4
4x2
x
Answer: (f ◦ g)(x) = 4+x
=
2
x(4 + x2 )
x2
8) Given f (x) =
x2
2−x
, find f −1 (x).
3x − 4
2 + 4x
Answer: f −1 (x) =
1 + 3x
9) Given f (x) =
10) Express the circumference of a circle as a function of its diameter d.
Answer: C = 2πr so C = πd, since d = 2r.
1
11) Find the real zero(s) of the function f (x) = 4 − (5 −
Answer: x =
11
4
or x =
√
4x − 2)2 .
51
4 .
12) The points (5, 2) and (4, −3) are on the graph of the function y = f (x).
Find the corresponding points on the graph obtained by shifting the graph
of f 2 units to the right, then reflecting it about the x-axis.
Answer: (7, −2) and (6, 3)
13) Find the equation of the line passing through the origin and perpendicular
to every line with slope 13 .
Answer: y = −3x
14) If f (x) =
domain.
x−3
and g(x) = x2 − 4, find the function f ◦ g and give its
2x
Answer: x 6= ±2 or x in the interval (−∞, −2) ∪ (−2, 2) ∪ (2, ∞)
15) Show that the function f (x) =
domain.
5
is 1-1. Find f −1 along with its
3 − 2x
Answer: Assume f (x1 ) = f (x2 ), so that
5
5
=
.
3 − 2x1
3 − 2x2
Cross-multiplying by the denominators yields
5(3 − 2x2 ) = 5(3 − 2x1 )
dividing by 5 yields
3 − 2x2 = 3 − 2x1
subtracting 3 yields
− 2x2 = −2x2
dividing by -2 yields
x2 = x1 .
Thus, the function f is one-to-one.
16) The graphs of odd functions are symmetric about the origin. (True/False)
Answer: True
17) Determine the vertex of the quadratic function y = 2x2 − 4x + 7.
Answer: y = 2(x − 1)2 + 5 so the vertex is (1, 5)
18) How does the domain of
f (x)
g(x)
relate to the domains of f (x) and g(x)?
Answer:
If dom(f ) and dom(g) represent the domains of f and g,
respectively, then the
f
dom( ) = {x in dom(f ) ∩ dom(g)|g(x) 6= 0}.
g
In words, the domain of fg is the intersection of the domains of f and g,
except for those values of x where g(x) = 0.
2
19) The function f (x) =
mining f −1 .
4
is 1-1. Find the range of f −1 without deter2 − 7x
Answer: range(f −1 ) = dom(f ) = (−∞, 27 ) ∪ ( 27 , ∞)
5
20) Determine where the functions f (x) = 3(4 − x) and g(x) = −2x +
2
intersect.
33
Answer: intersect at ( 19
2 ,− 2 )
√
21) Given f (x) = x − 3, which is 1-1, sketch the graphs of f and f −1 without
determining f −1 .
√
Answer: The graph of f is the graph of y = x shifted to the right by 3
units. The graph of the inverse function is the graph of f reflected about
the line y = x.
y
f −1 (x)
y=x
f (x) =
√
x
(0, 0)
22) Find the domain of the function f (x) =
(x2
2x
√
.
− 1) 1 − 3x
Answer: (−∞, −1) ∪ (−1, 31 ]
23) Find the equation of the line that passes through (4, 7) and is parallel to
the line segment joining (1, 1) and (−2, 3).
Answer: y = − 32 x +
29
3
24) Find the point of intersection of the graphs of the functions f (x) = 3 − 5x
and g(x) = 3x + 1.
Answer: intersect at ( 41 , 74 )
3
x−3
25) Let f (x) =
√
3 − x and g(x) = x2 − 9. Find the domains of f g and
Answer: The domain of f g is (−∞, 3], and the domain of
(−3, 3).
f
g
f
g.
is (−∞, −3)∪
26) Let f (x) = −2x2 + 1, g(x) = 4x − 2 and h(x) = −3x. Find (h ◦ f ◦ g)(x).
Answer: (h ◦ f ◦ g)(x) = −3(−2(4x − 2)2 + 1) = 6(4x − 2)2 − 3.
−3x
is 1-1. Find the point on the graph of f −1
x−2
corresponding to the value x = 4 in the domain of f .
27) The function f (x) =
Answer: (−6, 4) is on the graph of f −1 since (4, −6) is on the graph of
f
1√
28) On the domain [−3, 0], f (x) =
9 − x2 is 1-1. Find the inverse of f and
4
give its domain.
√
Answer: f −1 (x) = − 9 − 16x2
29) Given the quadratic equation y = 3x2 − 18x + 24, complete the square
to yield y = f (x) in standard form.
Sketch the graph of f using graph
√
transformations applied to y = x.
Answer: y = 3(x − 3)2 − 3, so (h, k) = (3, −3) and a = 3. So, the graph
of y is the graph of x2 shifted right by 3, shifted down by 3, and stretched
vertically by 3. Since a > 0, the graph is facing upwards.
y = 3(x − 3)2 − 3
y
(0, 0)
x
4