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Problem 3: Determine whether the function f(x) is “odd” function, where :
$$f\left(x\right)={\mathrm{log}}_{e}\{x+\sqrt{\left({x}^{2}+1\right)}\}$$
Solution : In order to determine the nature of function with respect to even or odd, we check for f(-x). Here,
$$\Rightarrow f\left(-x\right)={\mathrm{log}}_{e}[-x+\sqrt{\{}{\left(-x\right)}^{2}+1\}]={\mathrm{log}}_{e}\{-x+\sqrt{\left({x}^{2}+1\right)}\}$$
The expression on the right hand side can not be explicitly interpreted whether it equals to f(x) or not. Therefore, we rationalize the expression of logarithmic function,
$$\Rightarrow f\left(-x\right)={\mathrm{log}}_{e}\left[\frac{\{-x+\sqrt{\left({x}^{2}+1\right)}\}X\{x+\sqrt{\left({x}^{2}+1\right)}\}}{\{x+\sqrt{\left({x}^{2}+1\right)}\}}\right]={\mathrm{log}}_{e}\left[\frac{-{x}^{2}+{x}^{2}+1}{\{x+\sqrt{\left({x}^{2}+1\right)}\}}\right]$$
$$\Rightarrow f\left(-x\right)={\mathrm{log}}_{e}1-{\mathrm{log}}_{e}\{x+\sqrt{\left({x}^{2}+1\right)}\}=-{\mathrm{log}}_{e}\{x+\sqrt{\left({x}^{2}+1\right)}\}=-f\left(x\right)$$
Hence, given function is an “odd” function.
Problem 4: Determine whether sinx + cosx is an even or odd function?
Solution : In order to check the nature of the function, we evaluate f(-x),
$$f\left(-x\right)=\mathrm{sin}\left(-x\right)+\mathrm{cos}\left(-x\right)=-\mathrm{sin}x+\mathrm{cos}x$$
The resulting function is neither equal to f(x) nor equal to “-f(x)”. Hence, the given function is neither an even nor an odd function.
It is easy to find the nature of function resulting from mathematical operations, provided we know the nature of operand functions. As already discussed, we check for following possibilities :
Based on above algorithm, we can determine the nature of resulting function. For example, let us determine the nature of "fog" function when “f” is an even and “g” is an odd function. By definition,
$$fog\left(-x\right)=f\left(g\left(-x\right)\right)$$
But, “g” is an odd function. Hence,
$$\Rightarrow g\left(-x\right)=-g\left(x\right)$$
Combining two equations,
$$\Rightarrow fog\left(-x\right)=f\left(-g\left(x\right)\right)$$
It is given that “f” is even function. Therefore, f(-x) = f(x). Hence,
$$\Rightarrow fog\left(-x\right)==f\left(-g\left(x\right)\right)=f\left(g\left(x\right)\right)=fog\left(x\right)$$
Therefore, resulting “fog” function is even function.
The nature of resulting function subsequent to various mathematical operations is tabulated here for reference :
------------------------------------------------------------------------------------
f(x) g(x) f(x) ± g(x) f(x) g(x) f(x)/g(x), g(x)≠0 fog(x)------------------------------------------------------------------------------------
odd odd odd even even oddodd even Neither odd odd even
even even even even even even------------------------------------------------------------------------------------
We should emphasize here that we need not memorize this table. We can always carry out particular operation and determine whether a particular operation results in even, odd or neither of two function types. We shall work with a division operation here to illustrate the point. Let f(x) and g(x) be even and odd functions respectively. Let h(x) = f(x)/g(x). We now substitute “x” by “-x”,
$$\Rightarrow h\left(-x\right)=\frac{f\left(-x\right)}{g\left(-x\right)}$$
But f(x) is an even function. Hence, f(-x) = f(x). Further as g(x) is an odd function, g(-x) = - g(x).
$$\Rightarrow h\left(-x\right)=\frac{f\left(x\right)}{-g\left(x\right)}=-h\left(x\right)$$
Thus, the division, here, results in an odd function.
There is an useful parallel here to remember the results of multiplication and division operations. If we consider even as "plus (+)" and odd as "minus (-)", then the resulting function is same as that resulting from multiplication or division of plus and minus numbers. Product of even (plus) and odd (minus) is minus(odd). Product of odd (minus) and odd (minus) is plus (even). Similarly, division of odd (minus) by even (plus) is minus (odd) and so on.
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