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We, now, define valid operations for real functions in the light of discussion about the domain in the previous section. For different function operations, let us consider two real functions “f” and “g” with domains “ ${D}_{1}$ ” and “ ${D}_{2}$ ” respectively. Clearly, these domains are real number set R or subsets of R :
$${D}_{\mathrm{1,}}{D}_{2}\in R$$
The addition of two real functions is denoted as “f+g”. It is defined as :
$$f+g:{D}_{1}\cap {D}_{2}\to R\phantom{\rule{1em}{0ex}}\text{such that :}$$
$$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in {D}_{1}\cap {D}_{2}$$
The subtraction of two real functions is denoted as “f-g”. It is defined as :
$$f-g:{D}_{1}\cap {D}_{2}\to R\phantom{\rule{1em}{0ex}}\text{such that :}$$
$$\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in {D}_{1}\cap {D}_{2}$$
Scalar, here, means a real number constant, say “a”. The scalar multiplication of a real function with a constant is denoted as “af”. It is defined as :
$$af:{D}_{1}\to R\phantom{\rule{1em}{0ex}}\text{such that :}$$
$$\left(af\right)\left(x\right)=af\left(x\right)\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in {D}_{1}$$
The product of two real functions is denoted as “fg”. It is defined as :
$$fg:{D}_{1}\cap {D}_{2}\to R\phantom{\rule{1em}{0ex}}\text{such that :}$$
$$\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in {D}_{1}\cap {D}_{2}$$
The quotient of two real functions is denoted as “f/g”. It involves rational form as “f(x)/g(x)”, which is defined for g(x) ≠ 0. We need to exclude value of “x” for which g(x) is zero. Hence, it is defined as :
$$\frac{f}{g}:{D}_{1}\cap {D}_{2}-\left\{x\right|g\left(x\right)\ne 0\}\to R\phantom{\rule{1em}{0ex}}\text{suchthat :}$$
$$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in {D}_{1}\cap {D}_{2}-\left\{x\right|g\left(x\right)\ne 0\}$$
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 :
$$fg\left(x\right)=\sqrt{x}\left({x}^{2}-5x+6\right);\phantom{\rule{1em}{0ex}}x\ge 0$$
Similarly, quotient function, in rule form, is given as :
$$fg\left(x\right)=\frac{\sqrt{x}}{{x}^{2}-5x+6};\phantom{\rule{1em}{0ex}}x\ge \mathrm{0,}x\ne \mathrm{2,}x\ne 3$$
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 $\sqrt{(1-x)}$ ,
$$x-1\ge 0\phantom{\rule{1em}{0ex}}\Rightarrow x\ge 1$$ $$1-x\ge 0\phantom{\rule{1em}{0ex}}\Rightarrow x\le 1$$
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,
$$\text{Domain}=\left\{1\right\}$$
Problem : Find the domain of the function given by :
$$f\left(x\right)=\frac{x}{\sqrt{\left({x}^{2}-5x+6\right)}}$$
Solution :
The function is in rational form. We can treat numerator and denominator functions separately as f(x) and g(x). The numerator is valid for all real values of “x”. Hence, its domain is “R”.
$${D}_{1}=R$$
For determining domain of g(x), we are required to find the value of “x” for which square root in the denominator is real and not equal to zero. Thus, we need to evaluate square root expression for positive number. It means that :
$${x}^{2}-5x+6>0\phantom{\rule{1em}{0ex}}\Rightarrow \left(x-2\right)\left(x-3\right)>0$$
The roots of the corresponding quadratic equation is 2,3. Further, coefficient of " ${x}^{2}$ " term is a positive number (1>0) . Therefore, intervals on the sides are positive for the quadratic expression. The valid interval satisfying the inequality is :
$$x<2\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}x>3$$
$${D}_{2}=\left(-\infty ,2\right)\phantom{\rule{1em}{0ex}}\cup \phantom{\rule{1em}{0ex}}\left(\mathrm{3,}\infty \right)$$
Now, given function is quotient of two functions. Hence, domain of the given function is intersection of two domain excluding interval that renders denominator zero. However, we have already taken into account of this condition, while determining domain of the function in the denominator. Hence,
$$\text{Domain}={D}_{1}\phantom{\rule{1em}{0ex}}\cup \phantom{\rule{1em}{0ex}}{D}_{2}=R\phantom{\rule{1em}{0ex}}\cap \phantom{\rule{1em}{0ex}}\{\left(-\infty ,2\right)\phantom{\rule{1em}{0ex}}\cup \phantom{\rule{1em}{0ex}}\left(\mathrm{3,}\infty \right)\}$$
$$\Rightarrow \text{Domain}=\{\left(-\infty ,2\right)\phantom{\rule{1em}{0ex}}\cup \phantom{\rule{1em}{0ex}}\left(\mathrm{3,}\infty \right)\}$$
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