Let f(x) be a function.
To find whether f(x) is even or odd, we have to replace 'x' by '-x' in f(x). We have to conclude f(x) as even or odd function from the result of f(-x) as shown below.
1. If f(-x) = f(x), then f(x) is even function
2. If f(-x) = - f(x), then f(x) is odd function
If f(-x) is neither equal to f(x) nor -f(x), we have to conclude f(x) is neither even nor odd.
The sum or difference of two even functions is always even.
The sum or difference of two odd functions is always odd.
Even though we have many applications of even and odd functions, let us consider the important application of even and odd functions in integral calculus.
If f(x) is even function,
If f(x) is odd function,
Example 1 :
Let f(x) = x^{3}, is f(x) odd or even function ?
Solution :
To know f(x) is odd or even function, let us plug x = -x in f(x).
Then, we have
f(-x) = (-x)^{3}
f(-x) = -x^{3}
f(-x) = - f(x)
So, f(x) is odd function.
Example 2 :
Let f(x) = x^{2} + 2, is f(x) odd or even function ?
Solution :
To know f(x) is odd or even function, substitute -x for x in f(x).
Then, we have
f(-x) = (-x)^{2} + 2
f(-x) = x^{2} + 2
f(-x) = f(x)
So, f(x) is even function.
Example 3 :
Let f(x) = x^{3} - 2x, is f(x) odd or even function ?
Solution :
To know f(x) is odd or even function, substitute -x for x in f(x).
Then, we have
f(-x) = (-x)^{3} - 2(-x)
f(-x) = -x^{3} + 2x
f(-x) = -(x^{3} - 2x)
f(-x) = -f(x)
So, f(x) is odd function.
Example 4 :
Let f(x) = 5x^{3} + x^{2 }- 1, is f(x) odd or even function ?
Solution :
To know f(x) is odd or even function, substitute -x for x in f(x).
Then, we have
f(-x) = 5(-x)^{3} + (-x)^{2 }- 1
f(-x) = 5(-x^{3}) - x^{2 }- 1
f(-x) = -5x^{3} - x^{2 }- 1
f(-x) = -(5x^{3} + x^{2 }+ 1)
f(-x) can not be expressed as either as f(x) or -f(x).
So, f(x) is neither even nor odd function.
Example 5 :
Let f(x) = x^{4} + 2x^{2} + 5, is f(x) odd or even function ?
Solution :
To know f(x) is odd or even function, substitute -x for x in f(x).
Then, we have
f(-x) = (-x)^{4 }+ 2(-x)^{2} + 2
f(-x) = x^{4 }+ 2x^{2} + 2
f(-x) = f(x)
So, f(x) is even function.
Important Note :
In trigonometric ratios, if we have negative angle, we have to understand that the angle will fall in the IV^{th }quadrant.
In IV^{th} quadrant, the trigonometric ratios 'cos' and 'sec' are positive and all other trigonometric ratios are negative
Example 6 :
Is sinx odd or even function ?
Solution :
Let f(x) = sinx
To know f(x) is odd or even function, substitute -x for x in f(x).
Then, we have
f(-x) = sin(-x)
Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant "sin" is negative.
So, we have
f(-x) = - sinx
f(-x) = - f(x)
f(x) is odd function
So, sinx is odd function.
Example 7 :
Is cosx odd or even function ?
Solution :
To know f(x) is odd or even function, substitute -x for x in f(x).
Then, we have
f(-x) = cos(-x)
Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant "cos" is positive.
So, we have
f(-x) = cosx
f(-x) = f(x)
f(x) is even function
So, cosx is even function.
Example 8 :
Is tanx odd or even function ?
Solution :
Let f(x) = tanx
To know f(x) is odd or even function, substitute -x for x in f(x).
Then, we have
f(-x) = tan(-x)
Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant "tan" is negative.
So, we have
f(-x) = - tanx
f(-x) = - f(x)
f(x) is odd function
So, tanx is odd function.
Example 9 :
Let f(x) = sinx + tanx, is f(x) odd or even function ?
Solution :
To know f(x) is odd or even function, substitute -x for x in f(x).
Then, we have
f(-x) = sin (-x) + tan(-x)
Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant both "sin" and "tan" are negative.
So, we have
f(-x) = - sinx - tanx
f(-x) = - (sinx + tanx)
f(-x) = - f(x)
So, f(x) is odd function.
Note : The sum or difference of two odd functions is always odd.
Example 10 :
Let f(x) = secx + cosx, is f(x) odd or even function ?
Solution :
To know f(x) is odd or even function, substitute -x for x in f(x).
Then, we have
f(-x) = sec(-x) + cos(-x)
Because the angle is negative, it falls in the IV^{th} quadrant. In IV^{th} quadrant both "sec" and "cos" are positive.
So, we have
f(-x) = secx + cosx
f(-x) = f(x)
So, f(x) is even function.
Note : The sum or difference of two even functions is always even.
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