Math 122 – Midterm III Review
Types of Questions to Study – Solutions
1. 1. Which of the following are functions?
a.
d.
f.
x
1
2
-3
-1
0
y
1
4
9
1
0
b.
x y
0 0
1 1
4 2
1 -1
9 -3
c.
x
0
1
3
4
2
y
5
5
5
5
5
e.
g.
a, c, e, f are all functions. They either pass the vertical line test, or have a unique output
for every input.
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2. Use the graph of the function f below to answer the following:
a) What is f (1) ?
f (1) = 2
b) What is f (2) ?
f (2) = 4
c) What is f (6) ?
No solution, Undefined, Undetermined, DNE.
d) For what value or values of x is f (x) = 4 ?
f (x) = 4 whenever x = 0, 2, 4.
e) For what values of x is f increasing?
f is increasing when x ∈ [1, 3].
f) For what values of x is f decreasing?
f is decreasing when x ∈ [0, 1] ∪ [3, 5].
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g) What is the average rate of change of f over the interval [1, 3] ?
AV EROC
=
f (3) − f (1)
3−1
=
6−2
3−1
=
4
2
= 2
h) Find the local maxima and minima of f
The local maxima happen at x = 0 and x = 3, or at the points (0, 4) and (3, 6).
The local minima happen at x = 1 and x = 5, or at the points (1, 2) and (5, 1).
i) State the domain and range of f
Df : [0, 5]
Rf : [1, 6]
3. Let r(x) = x2 + x.
a) Find the average rate of change of r from (−3, r(−3)) to (4, r(4))
AV EROC
=
=
r(4) − r(−3)
4 − (−3)
2
(4) + 4 − (−3)2 + (−3)
7
=
20 − 6
7
=
14
7
= 2
4
b) Find the average rate of change of r from (a, r(a)) to (a + h, r(a + h))
r(a + h) − r(a)
(a + h) − (a)
(a + h)2 + (a + h) − a2 + a
h
2
a + 2ah + h2 + a + h − a2 + a
h
=
=
=
2ah + h2 + h
h
=
h(2a + h + 1)
h
= 2a + h + 1
4. Let
f (x) =
1−x
2x + 1
x≤1
x>1
a) Sketch the graph of f .
b) Evaluate f (−2) and f (1)
f (−2) = 3
f (1) = 0
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5. If f (x) = x2 + 1 and g(x) = x − 3, find the following.
a) f ◦ g
f ◦ g = f (g(x))
= f (x − 3)
= (x − 3)2 + 1
= x2 − 6x + 9 + 1
= x2 − 6x + 10
b) g ◦ f
g◦f
= g(f (x))
= g x2 + 1
= x2 + 1 − 3
= x2 − 2
c) f (g(2))
f (g(2)) = f ((2) − 3)
= f (−1)
= (−1)2 + 1
= 1+1
= 2
d) g(f (a + h))
g(f (a + h)) = g (a + h)2 + 1
= (a + h)2 + 1 − 3
= a2 + 2ah + h2 − 2
e) g ◦ g ◦ g
g◦g◦g =
=
=
=
=
=
g(g(g(x)))
g(g(x − 3))
g((x − 3) − 3)
g(x − 6)
(x − 6) − 3
x−9
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f) Is f an even or odd function? Explain.
f is an even function, as it is symmetric about the y-axis. This is because we know the
graph of y = x2 is symmetric about the y-axis, and y = x2 + 1 is the graph of x2 shifted
up 1, so it will also be symmetric about the y-axis, which makes it an even function.
On the other hand, we can conclude this by showing f (−x) = f (x).
f (−x) = (−x)2 + 1
= x2 + 1
Since f (−x) = f (x), f must be an even function.
g) Is g an even or odd function? Explain.
g is neither even nor odd. It is not symmetric about the y-axis, as no line is. And it
is not symmetric about the origin. This is easier shown algebraically. We can show that
g(−x) 6= −g(x) (odd) and g(−x) 6= g(x) (even).
g(−x) = (−x) − 3
= −x − 3
Since −g(x) = −(x − 3) = −x + 3, and g(x) = x − 3, and neither are equal to g(−x) =
−x − 3, we conclude that g is neither even nor odd.
6. Use the table below to answer the following:
x
0
a
b
c
p(x)
0
b
c
a
q(x)
a
c
b
d
a) p(q(a)) = p(c) = a
b) p(p(c)) = p(a) = b
c) q(p(0)) = q(0) = a
d) p(q(c)) = p(d) Undetermined/DNE/No Sol’n/Undefined
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7. a) How is the graph of y = f (x − 3) + 2 obtained from the graph of f ?
y is obtained by shifting f to the right 3 units, and then up 2 units.
b) How is the graph of y = f (−x) obtained from the graph of f ?
y is obtained by reflecting f over the y-axis.
c) How is the graph of y = −f (2x) − 5 obtained from the graph of f ?
y is obtained by horizontally shrinking f by a factor of 2 (cutting each x value in half),
reflecting it over the x-axis, and down 5 units.
8. The graph of the function y = j(x) is given below.
Match the equations (i) − (iv) with one of the graphs below.
(i) y = j(x + 2)
(ii) y = j(x) + 2
(i) - (c) j shifted to the left 2 units.
(ii) - (e) j shifted up 2 units.
(iii) - (d) j shifted down 2 units.
(iv) - (a) j reflected over the x-axis.
(iii) y = j(x) − 2
(iv) y = −j(x)
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9. Let g(x) = 3x4 − 14x2 + 5x − 3
a) Draw a rough sketch of g
b) Estimate the values of the local maximum and minimum values of g
There is one local maximum, when x ≈ 0.2, or at approximately (0.2, −2.5)
c) Estimate the interval(s) on which g is increasing and on which g is decreasing
g is increasing approximately when x ∈ [−1.65, 0.2] ∪ [1.45, ∞)
g is decreasing approximately when x ∈ (−∞, −1.65] ∪ [0.2, 1.45]
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10. Let m(x) =
√
x + 1 and n(x) = x − 1.
a) What is the domain of m?
We need x + 1 ≥ 0 so this means x ≥ −1. Thus Dm : [−1, ∞).
b) What is the domain of n?
Dn : (−∞, ∞)
c) What is the domain of m + n?
Dm+n = Dm ∩ Dn , so Dm+n : [−1, ∞)
d) What is the domain of m(n)?
p
√
m(n(x)) = m(x − 1) = (x − 1) + 1 = x, so it seems as if x ≥ 0. We must also
double check the domain of the inside function, n(x), and since the domain of n is all real
numbers, we have nothing else to worry about. So Dm(n) : [0, ∞).
e) What is the domain of n(m)?
√
√
n(m(x) = n x + 1 = x + 1 − 1, so it seems as if x ≥ −1. Again we double check the
inside function, and since m has the same domain, we can conclude that Dn(m(x)) : [−1, ∞)
f) What is the domain of m/n?
We can start with looking at Dm ∩ Dn , which is [−1, ∞), but we must also look at the
function in the denominator, because it is a fraction. So,
m √
= x + 1x − 1
n
From the denominator, we see that x 6= 1, or we would divide by zero, so we need to
also exclude x = 1 from our domain, giving us
Dm/n : [−1, 1) ∪ (1, ∞)
g) What is the domain of n/m?
x−1
n
=√
m
x+1
Since x 6= −1, or we would divide by zero, we have to exclude this value as well. So
Dn/m : (−1, ∞).
We first again start with [−1, ∞), but we also consider that
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11. A publisher estimates that the cost C(x) of printing a run of x copies of a certain
mathematics textbook is given by the function C(x) = 5000 + 30x − 0.001x2
a) Find C(1000) and C(10, 000)
C(1000) = 5000 + 30(1000) − 0.001(1000)2 = $34, 000
C(10, 000) = 5000 + 30(10, 000) − 0.001(10, 000)2 = $205, 000
b) What do your answers in part (a) represent?
The answers in part (a) represent the cost of printing 1000 copies of the textbook, and
the cost of printing 10, 000 copies of the textbook.
c) Find C(0). What does this number represent?
C(0) = 5000 + 30(0) − 0.001(0)2 = 5000
This is the cost of printing 0 textbooks, thus it must be the initial base fare to get any
copies printed at all.
12. Write an equation that expresses the relationship between the variables:
a) M varies directly as z
M = kz
b) z is inversely proportional to y
z=
k
y
c) F is jointly proportional to q1 and q2
F = kq1 q2
d) R is directly proportional to the square of v
R = kv 2
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13. The intensity of illumination I from a light varies inversely as the square of the distance
d from the light.
a) Write this statement as an equation.
I=
k
d2
b) Determine the constant of proportionality if it is known that a lamp has an intensity
of 1000 candles at a distance of 8 m.
I =
k
d2
1000 =
k
82
1000 =
k
64
64, 000 = k
So I =
64, 000
.
d2
c) What is the intensity of this lamp at a distance of 20 m?
I =
64, 000
d2
=
64, 000
(20)2
=
64, 000
400
I = 16, 000 candles
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12. The maximum weight M that can be supported by a beam is jointly proportional to
its width w and the square of its height h and inversely proportional to its length L.
a) Write an equation that expresses this proportionality.
M =k
wh2
L
b) Determine the constant of proportionality if a beam 4 in. wide, 6 in. high, and 12 ft.
long can support a weight of 4800 lb.
M
= k
wh2
L
4800 = k
(4)(6)2
12
4800 = k
144
12
4800 = k(12)
400 = k
So M = 400
wh2
L
c) If a 10-ft beam made of the same material is 3 in. wide and 10 in. high, what is the
maximum weight it can support?
M
wh2
L
(3)(10)2
= 400
10
300
= 400
10
= 400
= 400(30)
M
= 12, 000 lbs.