3.17 Composition of functions (exercise)  (Page 29/2)

Working rules

A. Writing composition :

• Write fog(x) = f(y), where y = g(x)

B. Finding domain of the composition :

• Write down the domain interval of argument function “g(x)” of the composition.
• Write the domain interval of the main function “f(x)” by substituting independent variable by the argument function “g(x)” itself.
• Interpret the interval with argument function, “g(x)”.
• Intersection of two intervals is the valid interval of composition.

Problem 1: A function f(x) is given as :

$$f\left(x\right)=\{a-x^n{\}}^{\frac{1}{n}}$$

where a>0, x>0 and "n" is a positive integer. Find f{f(x)}

Solution :

Statement of the problem : The domain of the given function is positive number as x>0. In order to find, the composition, we evaluate f(y), where y = f(x).

$$\Rightarrow f\left\{f\left(x\right)\right\}=f\left(y\right)={\left(a-y^n\right)}^{\frac{1}{n}}=[a-\{{\left(a-x^n\right)}^{1/n}{\}}^n{]}^{1/n}$$

$$\Rightarrow f\left\{f\left(x\right)\right\}={\left(a-a-x^n\right)}^{\frac{1}{n}}={\left(x^n\right)}^{\frac{1}{n}}=x;x>0$$

Here, composition is that of function with itself. As such, domain of composition is equal to intersection of domain of the given function with itself. But, the intersection of an interval with itself is same interval. Hence, we have retained the domain interval of the composition same as that of given function.

Problem 2: A function f(x) is given as :

$$f\left(x\right)=\{2-x^n{\}}^{\frac{1}{n}}$$

where x>0 and "n" is a positive integer. Prove that :

$$f\left\{f\left(x\right)\right\}+f\left\{f\left(\frac{1}{x}\right)\right\}\ge 2$$

Solution :

Statement of the problem : The domain of the given function is positive number as x>0. In order to prove the inequality, we need to determine each composition on the left hand side of the given inequality.

We have seen in earlier example that if $f\left(x\right)=\{a-x^n{\}}^{1∕n}$ , then f{f(x) = x. Hence, if $f\left(x\right)=\{2-x^n{\}}^{1∕n}$ , then f{f(x) = x. Similarly, we determine f{f(1/x)}. Here,

$$\Rightarrow f\left\{f\left(\frac{1}{x}\right)\right\}=f\left(y\right)=[a-\{{\left(a-\frac{1}{x^n}\right)}^{1/n}{\}}^n{]}^{1/n}={\left(a-a-\frac{1}{x^n}\right)}^{1/n}$$

$$\Rightarrow f\left\{f\left(\frac{1}{x}\right)\right\}={\left(\frac{1}{x^n}\right)}^{\frac{1}{n}}=\frac{1}{x}$$

Substituting these values in the LHS of the inequality, we have :

$$\Rightarrow \text{LHS}=f\left\{f\left(x\right)\right\}+f\left\{f\left(\frac{1}{x}\right)\right\}=x+\frac{1}{x}$$

Using algebraic identity $a^2+b^2={\left(a-b\right)}^2+2ab$ , we have :

$$\Rightarrow \text{LHS}={\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)}^2+2$$

But, the square term is a non-negative number. Hence,

$$\Rightarrow \text{LHS}\ge 2$$

$$\Rightarrow f\left\{f\left(x\right)\right\}+f\left\{f\left(1/x\right)\right\}\ge 2$$

Problem 3: A function is defined as :

| -1 ; -2≤x≤0 f(x) = || x-1 ; 0≤x≤2

Find composition f(|x|) and its domain.

Solution :

Statement of the problem : The function is defined by different rules in two intervals.

The composition consists of two functions “f(x)” and “|x|”. We know that modulus is defined for all values of “x”. However, domain of “f(x)” is [-2,2]. Hence, domain of composition is intersection of two domains, which is [-2,2]. Here,

| -1 ; -2≤|x|≤0 and R f(|x|) = || |x|-1 ; 0≤|x|≤2 and R

The interval “-2≤[|x|≤0” can be interpreted in parts. The left part is “|x|≥-2”, which is always true. The right part “|x|≤0” is meaningless, which yields no solution for “x”. Therefore, upper interval of the function is not a valid interval. On the other hand, interval “0≤|x|≤2” has two parts. The left part “|x|≥0” is true for all values of “x”. The right part is “|x|≤2”. This expands to :

$$-2\le x\le 2$$

The intersection of “-2≤x≤2” and “R” is “-2≤x≤2”. Hence, composition is :