Even and odd Nature of Functions
EVEN / ODD FUNCTIONS
A function is even if \(f(x) = f( - x).\) This means that the function is the same for \(\text{+ve } x\text{-axis}\) and \(\text{-ve } x\text{-axis,}\) or graphically, symmetric about the \(y\text{-axis}\)
For example, \(f(x) = {x^2}\) is even
A function is even about \('a'\) if it is symmetric about the line \(x=a.\) For example, \(f(x) = {(x - 1)^2}\) is even about \(x=1.\)
For such a function \(f(a - x) = f(a + x)\)
A function is odd if \(f(x)= f(-x),\) that is, the function on one side of \(x\text{-axis}\) is sign inverted with respect to the other side or graphically, symmetric about the origin.
For example, \(f(x) = {x^3}\) is odd.
As in the even case, \(f(x)\) can be odd about \('a'.\) For example, \(f(x) = {(x + 2)^3}\) is odd about \(x=-2.\)
A function can also be neither even nor odd.
For example, \(f(x)={x^2}+{x^3},\text{ as }f(x)\ne f(-x)\text{ and }f(x)\ne- f(-x)\)
Example - 34
Check whether the following functions are even or odd.
\(\begin {align} {\bf{(a)}} \ \; f(x) = \sin x - \cos x\qquad \qquad {\bf{(b)}} \ \; f(x) = \frac{x}{{{e^x} - 1}} + \frac{x}{2} + 1 \qquad \qquad {\bf{(c)}} \ \; f(x) = \sqrt[3]{{{{(1 - x)}^2}}} + \sqrt[3]{{{{(1 + x)}^2}}}\end {align}\)
Solution:
(a) \(f(-x)=\sin(- x)-\cos(-x)=-{\text{sin}}x-{\text{cos}}x\)
We see that \(f( - x) \ne f(x)\) and \(f( - x) \ne - f(x)\)
Hence, this function is neither even nor odd.
(b) \(f( - x) =\begin{align} \frac{{ - x}}{{{e^{ - x}} - 1}} - \frac{x}{2} + 1= \frac{{x{e^x}}}{{{e^x} - 1}} - \frac{x}{2} + 1\end{align}\)
\(= \begin{align}\frac{{x{e^x} + x}}{{2({e^x} - 1)}} + 1=\frac{{x({e^x} - 1) + 2x}}{{2({e^x} - 1)}} + 1=\frac{x}{{{e^x} - 1}} + \frac{x}{2} + 1 = f(x)\end{align}\)
Therefore, this function is even.
(c) \(f( - x) = f(x)\), which is obvious just by observation.
This function is even.
Example – 35
If the function \(f\) satisfies \(f(x + y) + f(x - y) = 2f(x).\,f(y)\) \(\forall x, y \; in\; \mathbb{R} \), and f(0) \( \ne\) 0, prove that \(f(x)\) is even
Solution:
Substituting \(x=y=0,\)
we get \(2f(0) = 2f{(0)^2}\) \( \Rightarrow \) \(f(0)\,\,(f(0) - 1) = 0\)
Since \(f(0)\) \( \ne\) 0, we have \(f(0)\) = 1
Now substitute \(x=0,\)
to get \(f(y) + f( - y) = 2f(0)f(y) = 2f(y)\)
\( \Rightarrow \) \(f(y) = f( - y)\)
\( \Rightarrow \) \(f\) is even.
Example – 36
If \(f(x + y) = f(x).f(y)\forall \) real \(x,\,y\) and \(f(0) \;\ne\,0,\) then prove that the function \(f(x) = \begin{align}\frac{{f(x)}}{{1 + {{\left( {f(x)} \right)}^2}}}\end{align}\) is an even function.
Solution:
As in the previous question, we first substitute \(x=y=0\) to get \(f(0)=1\) \(\Rightarrow \) \(f(x) =\begin{align} \frac{1}{{f( - x)}}\end{align}\)
(Note that the approach in such questions is always to try out some sample substitutions that yield significant results).
\[Now \qquad\qquad\qquad f( - x) = \frac{{f( - x)}}{{1 + {{\left( {f( - x)} \right)}^2}}} = \frac{{\frac{1}{{f(x)}}}}{{1 + {{\left( {\frac{1}{{f(x)}}} \right)}^2}}} = \frac{{f(x)}}{{1 + {{\left( {f(x)} \right)}^2}}} = f(x).\]
Hence, \(f(x)\) is even.