3.6 Function operations  (Page 2/2)

Problem : Let two functions be defined as :

$$f\left(x\right)=\sqrt{x}$$

$$g\left(x\right)=x^2-5x+6$$

Find domains of “fg” and “f/g”. Also define functions fg(x) and (f/g)(x).

Solution : The function f(x) is defined for all non negative real number. Hence, its domain is :

$$\Rightarrow x\ge 0$$

The function, g(x), - being a real quadratic polynomial - is real for all real values of “x”. Hence, its domain is “R”. Domains of two functions are shown in the figure.

The domain of fg(x) is intersection of two intervals, which is non-negative interval as shown in the figure :

$$D=D_1\cap D_2=\left[\mathrm{0,}\infty \right]\cap R$$

Here, we recall that intersection of a set with subset is equal to subset :

$$D=\left[\mathrm{0,}\infty \right]$$

This is the domain of product function “fg(x)”. The domain of quotient function “f/g(x)” excludes values of “x” for which “g(x)” is zero. In other words, we exclude roots of “g(x)” from domain. Now,

$$g\left(x\right)=x^2-5x+6=0$$

$$\Rightarrow g\left(x\right)=\left(x-2\right)\left(x-3\right)=0$$

$$\Rightarrow x=\mathrm{2,3}$$

Hence, domain of “f/g(x)” is :

$$\Rightarrow D_1\cap D_2-\left\{x\right|g\left(x\right)\ne 0\}=[\mathrm{0,}\infty ]-\{\mathrm{2,3}\}$$

Now the product function, in rule form, is given as :

$$fg\left(x\right)=\sqrt{x}\left(x^2-5x+6\right);x\ge 0$$

Similarly, quotient function, in rule form, is given as :

$$fg\left(x\right)=\frac{\sqrt{x}}{x^2-5x+6};x\ge \mathrm{0,}x\ne \mathrm{2,}x\ne 3$$

Problem : Find domain of the function :

$$f\left(x\right)=2\sqrt{(x-1)}+\sqrt{(1-x)}+\sqrt{(x_2+x+1)}$$

Solution : Given function can be considered to be addition of three separate function. We know that scalar multiplication of a function does not change domain. As such, domain of $2\sqrt{(x-1)}$ is same as that of $\sqrt{(x-1)}$ . For $\sqrt{(x-1)}$ and $\sqrt{(1-x)}$ ,

$$x-1\ge 0\Rightarrow x\ge 1$$ $$1-x\ge 0\Rightarrow x\le 1$$

Now, we use sign rule for third function :

$$x^2+x+1\ge 0$$

Here, coefficient of “ $x^2$ ” is positive and D is negative. Hence, function is positive for all real x. This means f(x)>0. This, in turn, means f(x)≥0. The domain of third function is R. Domain of given function is intersection of three domains. From figure, it is clear that only x=1 is common to three domains. Therefore,

$$\text{Domain}=\left\{1\right\}$$