Method 1.
We have
For #6 we must ask: What do we have to plug in to f(x) to
get 5 out? So set f(x) = 5 and
solve for x ; we have
Therefore 
Similarly, on #8 we set g(x) = 2 and solve for x:
Therefore 
Note. You will not be responsible for logarithms on
this exam . :-)
Method 2.
First we have
Using the first formula we get
Using the second formula we get 
Using the third formula we get 
Using the fourth formula we get 
For #6 we compute f -1(x) by the “switch y and x” trick discussed in
class. We have
So 
. Therefore
Similarly for #8 we have
So 
Note. You will not be responsible for logarithms on this exam. :-)
Graphs.
1. The graph of f (x) is shown at right.
On the same axes, sketch the graph of f(−x).
The graph of f(−x) is shown with dashed lines.
Notice that it is the (horizontal) reflection of
f(x) about the y-axis. |
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2. On the axes below, sketch the graph of f(x) = 3x+2.
The graph of f(x) = 3x+2 is obtained by shifting that of
g(x) = 3x to the left 2 units. Both graphs are shown at
right. |
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3. On the axes below, sketch the graph of h(x) = 2|x − 1| + 3.
There are two ways to do this problem:
(1) Transformations of |x|. Notice that if we perform the fol lowing
transformations ,
we will get h(x):
Therefore the graph looks like the picture at right.
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(2)
Piecewise function.
h(1) = 2(1)+1 = 3, so the vertex is at (1, 3), and
we get the graph shown. |
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4. The graph of g(x) is shown at right.
On the same axes, sketch the graph of g-1(x).
To get the graph of the inverse of a function, reflect the graph
about the line y = x. The line y = x is shown for reference. |
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Work and Answer. You must show all relevant work to
receive full credit.
1. Write 
as a piecewise function and graph the function. What is the domain
of f(x)?
Notice that the first case says “>” rather than “ ≥” since
f(x) is undefined at x = 2. In
fact, the domain of f(x) is {x | x ≠ 2}.
From the piecewise function, we can see that the graph looks like the one shown .
2. Find the inverse of the function
.
To compute the inverse of a one-to-one function, switch y and x, then solve for
the new
y. We have
Multiply both sides by 2y − 5:
x(2y − 5) = 3
Divide both sides by x:
3. Find the domain of the function
. Ex press your answer in interval notation.
Because of the denominator , 4x − 1 ≠ 0. Solving for x, we get 4x ≠ 1, or
.
We also have, because of the square root , x ≥ 0. So the domain is x ≥0 but
. In
interval notation this is 
