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Question: A function is of the form \(f\left( x \right) = ax + b\) . If \(\left( {fof} \right)\left( x \right)...

A function is of the form f(x)=ax+bf\left( x \right) = ax + b . If (fof)(x)=4x+9\left( {fof} \right)\left( x \right) = 4x + 9 , find the values of a and b.

Explanation

Solution

Let f:ABf:A \to B and g:BCg:B \to C be two functions. Then the composition of ff and gg , denoted by gofgof, is defined as the function gof:ACgof:A \to C given by gof(x)=g(f(x)),xAgof\left( x \right) = g\left( {f\left( x \right)} \right),\forall x \in A.

Clearly, gof(x)gof\left( x \right) is a function in which variables of the function g(x)g\left( x \right) are replaced by f(x)f\left( x \right) .To find (fof)(x)\left( {fof} \right)\left( x \right) replace x by ax+bax+b in the given function f(x)=ax+bf\left( x \right) = ax + b , we get an equation and equate it to the given value 4x+94x + 9 and compare the coefficients to get the value a and b.

Complete step-by-step answer:
Given f(x)=ax+bf\left( x \right) = ax + b
We have to find (fof)(x)\left( {fof} \right)\left( x \right).
Replace x by ax+bax+b and equate it to the value 4x+94x + 9
(fof)(x)=f(f(x))\left( {fof} \right)\left( x \right) = f\left( {f\left( x \right)} \right)
f(f(x))=f(ax+b)f\left( {f\left( x \right)} \right) = f\left( {ax + b} \right)

a(ax+b)+b\Rightarrow a\left( {ax + b} \right) + b

a2x+ab+b\Rightarrow {a^2}x + ab + b

Find values of a and b
It is given that (fof)(x)=4x+9\left( {fof} \right)\left( x \right) = 4x + 9
Hence, a2x+ab+b=4x+9{a^2}x + ab + b = 4x + 9
We can compare the coefficients of xx and the constant term on both sides because we are given that (fof)(x)\left( {fof} \right)\left( x \right) exactly equals to the 4x+94x + 9 for all values of xx , i.e. (fof)(x)=4x+9\left( {fof} \right)\left( x \right) = 4x + 9 is an identity.
Thus, we can compare the coefficients of various powers of xx.

a2=4{a^2} = 4

a=4\Rightarrow a = \sqrt 4

a=±2\because a = \pm 2

As 2 ×\times 2 = 4 and also (-2) ×\times (-2) = 4.

ab+b=9ab + b = 9

b(a+1)=9\Rightarrow b\left( {a + 1} \right) = 9

Put values of a to calculate values of b
For a = 2

b(a+1)=9b\left( {a + 1} \right) = 9

b(2+1)=9\Rightarrow b\left( {2 + 1} \right) = 9

b×3=9\Rightarrow b \times 3 = 9

b=93\Rightarrow b = \dfrac{9}{3}

b=3\because b = 3

For a = -2

b(a+1)=9b\left( {a + 1} \right) = 9

b(2+1)=9\Rightarrow b\left( { - 2 + 1} \right) = 9

b×(1)=9\Rightarrow b \times \left( { - 1} \right) = 9

b=9\because b = - 9

The values of (a, b) are (2, 3) and (-2, -9).

Note: Similarly, fog(x)fog\left( x \right) is a function in which variables of the function f(x)f\left( x \right) are replaced by g(x)g\left( x \right).
The composition of functions is also useful in checking the invertible function, hence finding the inverse of the function.
A function f:XYf:X \to Y is defined to be invertible, if there exists a function g:YXg:Y \to X such that gof=Ixgof = {I_x} and fog=Iyfog = {I_y}. The function gg is called the inverse of ff and is denoted by f1{f^{ - 1}} .
We should always be cleared between an equation and an identity. An equation has some solutions, it is not satisfied by all the values of the domain of the variable while identity is satisfied by all the values in the domain of the variable.