082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

62 Chapter 2 Engineering functions
Example2.5 If f (t) = 2t +3 andg(t) = t +1 writeexpressions for the compositions
2
(a) f(g(t))
(b) g(f(t))
( ) t +1
Solution (a) f(g(t)) = f
2
The ruledescribing the function f is: ‘double the inputand then add 3’. Hence,
So
( ) ( ) t +1 t +1
f = 2 +3=t+4
2 2
f(g(t))=t+4
(b) g(f(t)) =g(2t +3)
The ruleforgis: ‘add 1 tothe inputand then divide everything by 2’. So,
Hence
g(2t +3) = 2t+3+1
2
g(f(t))=t+2
Clearly f (g(t)) ≠g(f(t)).
=t+2
2.3.7 Inverseofafunction
Considerafunction f (x).Itcanbethoughtofasacceptinganinputx,andproducingan
output f (x). Suppose now that this output becomes the input to the functiong(x), and
the output fromg(x) isx, thatis
g(f(x))=x
We can think ofg(x) as undoing the work of f (x). Figure 2.10 illustrates this situation.
Theng(x) is the inverse of f, and is written as f −1 (x). Since f −1 (x) undoes the work
of f(x)wehave
f (f −1 (x)) = f −1 (f(x))=x
x
f
f (x)
g
g( f (x))=x
Figure2.10
The functiongisthe inverse of f.