You have probably already met the formula for changing temperatures in degrees Celsius, °C, to degrees Fahrenheit, °F . It is
F = C + 321
You can regard this as a 1 : 1 mapping and it can be transformed to give C in terms of F, as follows:
F − 32 = C (taking −32 from each side)
(F − 32) = C (multiplying by )
That is,
C = (F − 32)2
If we write equations 1 and 2 in our functions notation,
writing C for x in 1 and
F for x in 2 ,
we have
f : x → x + 32
g : x → (x − 32)
Worked Examples
Show that fg(x) = gf(x) = x for the functions above.
| fg(x) = f(g(x)) | = g(x) + 32 |
| = ⎛ ⎝(x − 32)⎞ ⎠ + 32 | |
| = x − 32 + 32 | |
| = x |
and
| gf(x) = g(f(x)) | = (f(x) − 32) |
| = ⎛ ⎝x + 32 − 32⎞ ⎠ | |
| = × x | |
| = x |
Note
If functions f and g are such that
we say that g is the inverse of f and denote this by
f−1 (x) = g(x) or f−1 = g
Similarly, f is the inverse of g, so
g−1 (x) = f(x) or g−1 = f
We can see how to find inverse functions in the next Worked Example.
If f(x) = + 2 (x ≠ 1), find its inverse function and state its domain.
We write y = + 2 and, as with the temperature conversion above, use algebraic manipulation to write x as a function of y.
Starting with
| y | = + 2 | take −2 from each side to give |
| y − 2 | = + 2 − 2 | |
| = |
Multiply both sides by (1 − x) to give
| (1 − x)(y − 2) | =(1 − x) × |
| (1 − x)(y − 2) | = 1 |
Now divide both sides by (y − 2) to give
| = | |
| 1 − x | = |
or, multiplying throughout by −1,
x − 1 =
and adding 1 to both sides, gives
x − 1 + 1 = + 1
or
x = 1 +
This is the inverse function, which we could write as
f−1 (y) : y → 1 +
Interchanging y and x (it is just a variable) gives
f−1 (x) : x → 1 +
or
f−1 (x) = 1 +
The domain of f−1 (x) is x ≠ 2 (as the function is not defined at x = 2).
Note
1.
We can check values of
f(x) = + 2 and f−1(x) = 1 +
For example,
f(2) = + 2 = + 2 = −1 + 2 = 1
and
f−1(1) = 1 + = 1 + = 1 + 1 = 2
Similarly,
f(4) = + 2 = − + 2 =
whilst
f−1 ⎛
⎝⎞
⎠ =
1 + =
1 + = 1 + 3 = 4
So we see that, for these values,
f−1f(x) = x
2.
In general,
f−1f(x) = f−1(f(x)) =
f−1⎛
⎝ + 2⎞
⎠,
using the f formula
= f−1⎛
⎝⎞
⎠
= f−1⎛
⎝⎞
⎠
= f−1⎛
⎝⎞
⎠,
and using the f−1 formula
= 1 +
= 1 +
= 1 +
= 1 +
= 1 − 1 + x
= x
So this proves that f and f−1 are inverse functions.
Two functions, g and h, are defined as
g : x → and
h : x → .
Calculate
the value of g(7)
Show me g(7) = =the value of x for which g(x) = 6.
Show me| 6 = ⇒ 6(x − 4) | = 2x + 3 |
| 6x − 24 | = 2x + 3 |
| 4x | = 27 |
| x | = |
Check
| g⎛ ⎝⎞ ⎠ | = |
| = | |
| = | |
| = 6 |
Write expressions for
hg(x)
Show me| hg(x) | = h(g(x)) |
| = | |
| = | |
| = |
g−1(x)
Show meTo find g−1(x) , we write
y =
and find x as a function of y; that is
y(x − 4) = 2x + 3
yx − 4y = 2x + 3
yx − 2x = 3 + 4y
x(y − 2) = 3 + 4y
x = ⇒ g−1(y) =
So
g−1(x) = (replacing y by x).
Exercises
