5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book May 8, 2018 13:46

60 STEP 4. Review the Knowledge You Need to Score High

Example 3

Given the graphs of functionsf(x) andg(x) in Figures 5.3-2 and 5.3-3,

3

f(x)

(^021)
x
y
1
2
3
Figure 5.3-2
3
g(x)
(^021)
x
y
1
2
3
Figure 5.3-3
find:
(a) (f+g)(1)
(b) (fg)(0)
(c)
(
f
g
)
(0)
(d) f(g(3))
Solutions
(a) (f+g)(1)= f(1)+g(1)= 3
(b) (fg)(0)=f(0)g(0)=3(0)= 0
(c)
(
f
g
)
(0)=
f(0)
g(0)

=

3

0

undefined

(d) f(g(3))=f(1)= 1

Inverse Functions

Given a functionf, the inverse of f(if it exists) is a functiongsuch that f(g(x))=xfor

everyxin the domain ofgandg(f(x))=xfor everyxin the domain of f. The function

gis written as f−^1. Thus, f(f−^1 (x))=xand f−^1 (f(x))=x. The graphs of f and f−^1 in

Figure 5.3-4 arereflectionsof each other in the liney=x. The point (a,b) is on the graph

offif and only if the point (b,a) is on the graph of f−^1.

y y = x

f

f−^1

x

(a, b)

(b, a)

Figure 5.3-4

A functionf isone-to-oneif for any two pointsx 1 andx 2 in the domain such thatx 1 =x 2 ,

thenf(x 1 )= f(x 2 ).