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Math 116.01 - Midterm 2 Solutions, version 1
(1) Find the domain and range of the function
f (x) = log2 (6 − 2x).
Since the outside part of this function is a logarithm, the range of this function is all real
numbers.
For the domain, recall that the domain of every logarithm function loga (x) is (0, ∞). So
the part of the formula inside the log must be positive. So
0 < 6 − 2x
2x < 6
x < 3.
So the domain of this function is (−∞, 3).
2
(2) Sketch a graph of the function
1
−1
3−x
using graphical operations. Label each step with the operation you did and the formula for
that step.
g(x) =
Since the formula for the function g(x) most resembles 1/x, we’ll start with a graph of
1/x.
Then we’ll replace x with x + 3, causing a shift to the left by 3 units:
Math 116.01
Next, a horizontal flip by replacing x with −x:
And finally, subtract 1, causing a shift down 1 unit:
3
4
(3) If a company sells its product for p dollars per unit, they can sell q = 10000 − 10p.
(a) Write revenue as a function of p.
Revenue is r = pq. So writing everything as a function of p, we get
r(p) = q(p) · p = (10000 − 10p) · p = 10000p − 10p2 .
(b) What price should the company choose to maximize revenue, and what is the maximum
revenue?
Since the revenue function is quadratic (and the coefficient on the squared term is
negative), the maximum happens at the vertex. We can find the vertex by completing
the square:
r(p) = −10p2 + 10000p
= −10(p2 − 1000p)
= −10(p2 − 1000p + 5002 − 5002 )
= −10((p − 500)2 − 5002 )
= −10(p − 500)2 + 10 · 5002
= −10(p − 500)2 + 2500000.
So the maximum revenue is $2500000, which occurs when the price is p = $500.
Math 116.01
5
(4) Solve the system
3x + 2y − z = 2
2x + 2y − z = 10
−4x + y + 2z = 6
Solving the first equation for z, we get
z = 3x + 2y − 2.
Substituting this into the other equations, we get
2x + 2y − (3x + 2y − 2) = 10
−4x + y + 2(2x + 2y − 2) = 6.
Simplifying, we get
−x = 8
2x + 5y = 10.
The first equation tells us immediately that x = −8. Putting this into the second equation,
we get
2(−8) + 5y = 10
⇒
y = 26
5 .
Now that we know x and y, we can find z, since we solved for it earlier:
78
z = 3(−8) + 2( 26
5 −2=− 5 .
So the system has a single solution,
x = −8,
y=
26
5 ,
z = − 78
5 .
6
(5) Solve these equations for x:
2
(a) 2(x −2x) − 1 = 7
2(x
2 −2x)
−1=7
(x2 −2x)
2
(x2 −2x)
log2 (2
=8
) = log2 (8)
x2 − 2x = 3
x2 − 2x − 3 = 0
(x − 3)(x + 1) = 0
x = −1, 3.
(b)
2
3
log3 (2x + 1) = 3
2
3
log3 (2x + 1) = 3
log3 (2x + 1) = 3 ·
3
2
=
9
2
39/2 = 2x + 1
39/2 − 1 = 2x
x = 12 (39/2 − 1)
≈ 69.648 . . .