Functions and relations answers
ClementRadcliffe, Contributor

We began the review of functions and relations last week. In today's lesson, we will share the solution to last week's homework.

Given that ƒ : x = 3x - 2

g : x = 2x + 5

Evaluate:

(i) g(-6)

(ii) ƒg (3)

Solution:

(i) Since g : x = 2x + 5, g(x) = 2x + 5

... g(-6) = (2 x -6) + 5

= -12 + 5 = -7.

... g(-6) = -7

ii) Remember now that for fg(x), g(x) replaces x in f(x)

As g(x) = 2(x) + 5 and f(x) = 3x - 2

... ƒg(x) = ƒ (2x + 5)

... ƒg(x) = 3 (2x + 5) - 2

= 6x + 15 - 2 = 6x + 13

... ƒg(3) = (6 x 3) + 13 = 18 +13

... ƒg(3) = 31

• If ƒ(x) = 2x - 1 and g(x) = 1/2 (x + 2),
  calculate

(i) ƒ(3)

(ii) gƒ(3).

Solution:

(i) As ƒ(x) = 2x - 1, then ƒ(3) = (3 x 2) -1 = 6 - 1 = 5.

... ƒ(3) = 5

(ii) Since ƒ(3) = 5, then gƒ(3) = g(5) .

Since g(x) = 1/2 (x + 2)

... g(5) = 1/2 (5 + 2) = 7/2

... gf(3) = 7/2.

Now that we have gone through the homework, our lesson will continue.

Inverse of a Function

If ƒ is the function defined as y = ax + b, then f-1, the inverse function expresses the variable x in terms of y.

Example:

y = ax + b

... ax = y - b

... x = y - b/a (x is expressed as a function of y)

Interchange x for y. (This is necessary, as y is always expressed as a function of x)

... y = x - b/a

... f^-1(x) = (x - b)/a or

... f^-1 = (x - b)/a

that is, the inverse of function ƒ, (f^-1), is (x - b)/a

Please note that this method should always end with the statement:

f^-1 (x) = (x - b)/a and NEVER y = (x - b)/a.

Given the function y = ax + b, some students express f^-1(x) as 1 by assuming that -1/ax + b is the power of ƒ as in indices. I am sure you will never make this error.

Example:

Given that ƒ(x) = 1/2 (x + 2). Calculate f^-1(x)

Since f(x) = 1/2 (x + 2)

...y = 1/2 (x + 2)

2y = x + 2

...x = 2y - 2

Interchanging x for y (Always remember this step; it must also be explicitly stated).

...y = 2x - 2

... f^-1(x) = 2x - 2

Please be sure that you are comfortable with the methods of cross-multiplication and changing the subject of a formula.

Inverse of a Composite Function

Given the functions y = ƒ(x) and y = g(x), then y = gƒ(x) is a composite function.

Since gf(x) is a function of x, the inverse is found by using the method outlined above.

Example:

Given the functions ƒ(x) = 3x and g(x) = x - 2, determine the functions:

(a) fg(x)

(b) fg -1(x)

Solution:

(a) As f(x) = 3x and g(x) = x - 2

... ƒg(x) = f(x - 2) = 3(x - 2)

... ƒg(x) = 3(x - 2)

(b) y = ƒg(x) = 3(x - 2)

... y = 3x - 6

... 3x = y + 6

... x = (y + 6)/3 Interchange x for y

... y =( x + 6)/3

... The inverse of ƒg(x) OR (fg)-1(x) is (x + 6)/3

Let us attempt another example:

Example

Given f(x) = x² and g(x) = 5x + 3, calculate

(i) f(-2)

(ii) gf(-2)

(iii)(g f) -1x

Solution

(i) Since f(x) = x² ... f(-2) = (-2)2 = 4.

Answer: f(-2) = 4.

(ii) As seen from above, f(-2) = 4

And since g(x) = 5x + 3 ... gf(-2) = g(4) = (5 x 4) + 3 = 23

... gf(-2)= 23

(iii) Given that f(x) = x² and g(x) = 5x + 3

then gf(x) = g(x²)

Since g(x) = 5x + 3 ... g(x²) = 5x² + 3

... gf(x) = 5x² + 3

(NB. If gf(x) = 5x² + 3, then gf(-2) = 5 x (-2)2 + 3 = 23 as above.)

In order to find the inverse, then let y = gf(x)

... y = 5x² + 3

... 5x² = y - 3

... x² = Square root of (y - 3)/5

... x =square root of (y - 3)/5

Interchanging x for y

... y = square root of (x - 3)/5

... (g f) -1x = square root of (x - 3)/5

Please do the following for homework.

f and g are functions defined as follows

ƒ : x = 3x - 5

g : x = 1/2 x

(a) Calculate the value of ƒ(-3)

(b) Write expressions for (i) ƒ^-1(x)

(ii) g^-1 (x)

(c) Hence, or otherwise, write an expression for (gf)^-1

Given that f : x = x + 3 and g : x = 2x

(a) Determine fg^-1(x) and g^-1 f^-1 (x)

(b) Hence evaluate fg^-1 (5 ) and g^-1 f^-1 (5)

Have a good week.

Clement Radcliffe is the principal of Glenmuir High School in May Pen.