Discuss the concept of even and odd functions graphically and analytically.
Even FunctionsThe graph of even functions are symmetric with respect to the y axis.To proove analytically that a given function f is even, we need to proove that: Odd FunctionsThe graph of even functions are symmetric with respect to the origin (0,0). To proove analytically that a given function f is odd, we need to proove that:Example 1The formulas of the even functions f, g, h and i are given (see graphs above). Show analytically that each of these functions satisfies the property of an even function: f(-x) = f(x)Solution to Example 1 The formulae of the four functions f, g, h and i are given with the graphs of these functions. (see top graph above). 1) f(x) = x2 f(-x) = (- x)2 = (-x)(-x) = x2 From the above, f(-x) = f(x), hence f(x) is an even function. 2) g(x) = - | x | + 2 g(-x) = - | - x | + 2 The absolute value function function | x | is an even function, therefore | - x | = | x | we simplify g(-x) to g(-x) = - | x | + 2 g(-x) = g(x), hence g(x) is an even function. 3) h(x) = 10 e- 0.2 x2 h(- x) = 10 e - 0.2 (- x) 2 = 10 e - 0.2 x 2 Since h(-x) = h(x), h(x) is an even function. 4) i(x) = 8 cos(x) i(- x) = 8 cos( - x) The cosine function is an even function, therefore cos(-x) = cos(x) which gives i(- x) = 8 cos( - x) = 8 cos(x) i(-x) = i(x) and therefore i(x) ia an even function. Example 2The formulas of the odd functions j, k, l and m are given (see graphs above). Show analytically that each of these functions satisfies the property of an odd function: f(- x) = - f(x)Solution to Example 2 The formulae of the four functions j, k, l and m are given with the graphs of these functions. (see top graph above). 1) j(x) = x3 j(-x) = (- x)3 = (-x)(-x)(-x) = - x3 From the above, j(-x) = - j(x), hence j(x) is an odd function. 2) k(x) = - x k(-x) = - (-x) = x From the above, k(-x) = - k(x), hence k(x) is an odd function. 3) l(x) = 6 sin(x) l(-x) = 6 sin( - x) The sine function is an odd function, therefore sin(-x) = - sin(x) which gives l(-x) = 6 sin( - x) = - 6 sin(x) l(-x) = - l(x) and therefore l(x) ia an odd function. 3) m(x) = ex - e-x m(- x) = e(- x) - e-(-x) Simplify = e- x - ex and rewrite as = - ( - e- x + ex ) m(-x) = - m(x) and therefore m(x) is an odd function. ExercisesVerify analytically whether each of these functions is even, odd or neither?1. f(x) = - 2 x2 + 4 2. g(x) = 3 | x | - x2 3. h(x) = 1 / x 4. i(x) = tan(x) 5. j(x) = x2 + 3 x Detailed solutions to Above Exercises 1. f(x) = - 2 x2 + 4 Let us find f(-x) = - 2 (-x)2 + 4 = - 2 x2 + 4 f(-x) is equal to f(x) so function f is even. 2. g(x) = 3 | x | - x2 g(-x) = 3 | - x | - (-x)2 = 3 | x | - x2 and is equal to g(x) hence function g is even. 3. h(x) = 1 / x h(-x) = 1 / (-x) = - 1 / x and is equal to - h(x), and therefore h is odd. 4. i(x) = tan(x) i(-x) = tan(-x) = - tan(x), function i is odd. 5. j(x) = x2 + 3 x j(-x) = (- x)2 + 3 (- x) = x2 - 3 x j(-x) is not equal to j(x) and j(-x) is not equal to - j(x). Therefore j(x) is neither an even nor an odd function. More References and LinksProperties of Trigonometric FunctionsApplications, Graphs, Domain and Range of Functions |