# ConcepTest * Chapter 1

```Section 1.8
Transforming Functions
Applied Calculus ,4/E, Deborah Hughes-Hallett
The graphs in Figure 1.11 (I)–(IV) are that of y = f(x).
Which could be a graph of cf(x)?
Figure 1.11
The graphs in Figure 1.11 (I)–(IV) are that of y = f(x).
Which could be a graph of f(x) − k?
Figure 1.11
The graphs in Figure 1.11 (I)–(IV) are that of y = f(x).
Which could be a graph of f(x − h)?
Figure 1.11
Let f and g have values given in the table.
x
−2
−1
0
1
2
f(x) g(x)
1 −1
0
1
−2 2
2
0
−1 −2
4. f(g(1)) = -2
5. f(g(0)) = -1
6. f(g(-1)) = 2
7. If f(g(x)) = 1, then x = 2
8. If f(g(x)) = 0, then x = -2
9. If g(f(x)) = 2, then x = -1
10. If g(f(x)) = -2, then x = 1
Let the graphs of f and g be as shown. Estimate the
values.
g(f(0)) ≈ 1
g(f(8)) ≈ 5
g(f(3)) ≈ 2
f(g(2)) ≈ 1
f(g(-1)) ≈ 0
Given the graphs of the functions g and f in Figures 1.13 and
1.14, which of the following is a graph of f(g(x))?
g(x)
f(x)
Given the graph of π(π₯),
sketch the graphs of
a. 2 + π(π₯) c. π( π₯ + 1)
b. 2π π₯
d. 1/π(π₯)
6
4
2
6
4
2
2
2
4
6
4
6
One of the graphs below shows the rate of ο¬ow, R, of blood from the heart in a
man who bicycles for twenty minutes, starting at t = 0 minutes. The other graph
shows the pressure, p, in the artery leading to a man’s lungs as a function of the
rate of ο¬ow of blood from the heart.
• Estimate π(π 10 ) and π(π 22 ).
• π(π 10 ) ≈ π(18) ≈ 23
• π(π 22 ) ≈ π(10) ≈ 17
• Explain what π(π 10 ) means in practical terms.
• After 10 minutes of bicycling, the pressure in the artery leading to
the man’s lungs is about 17 mm Hg.
x2
18. If f ( x) ο½ x ο« 1 and g( x) ο½ e , then what are
f ( g( x)) and g( f ( x))?
2
π(π(π₯)) =
2
2π₯
π
=
π π π₯
+1
=π
2 2
π₯
π
π₯ 2 +1
2
+1
=π
π₯ 2 +1
The graph in Figure 1.15 could be that of
1
(a) y ο½ ln x ο«
2
1
(b) y ο½ ln x ο­
2
1
(c) y ο½ ln( x ο« )
2
1
(d) y ο½ ln( x ο­ )
2
Figure 1.15
19. Sketch graphs of the following functions.
(a) y = −ln(4 + x)
(b) y = ln(4 − x)
(c) y = 3x−4
(d) y = 34−x
```