# Functions II Odd and Even Functions By Mr Porter.

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Functions II Odd and Even Functions By Mr Porter

Even Functions A function, f(x), is an even function if f(x) = f(-x), for all x in its domain. An even function has the important property, line symmetry with the y-axis as it axis. This allows us to plot the right hand- side of the y-axis, then to complete the sketch, draw the mirror image on the left-side of the y-axis. Y-axis Right-hand side of f(x) y = f(x) Y-axis Y-axis is an axis of symmetry y = f(x)

Testing for an Even Function To test whether a function is even or odd or neither, work with f(-x) and compare it to f(x). Usually, it is also better to use x = a and x = -a as the two values for comparison. Examples Show that f(x) = x 2 - 3 is an even function. Lets use numerical values x = 3 and x = -3. Evaluate f(3) in f(x); f(x) = x 2 - 3 f(3) = (3) 2 - 3 f(3) = 6 andf(x) = x 2 - 3 f(-3) = (-3) 2 - 3 f(-3) = 6 Now, f(-3) = f(3) Hence, f(x) = x 2 - 3 is an even function. But, is the same true for x = ±4, ±12,.. To be certain, you should test every value of x! This is not practical. A better method, is to use algebraic values of x, say x = ±a f(x) = x 2 - 3 f(a) = (a) 2 - 3 f(a) = a 2 - 3 andf(x) = x 2 - 3 f(-a) = (-a) 2 - 3 f(-a) = a 2 - 3 Now, f(-a) = f(a) Hence, f(x) = x 2 - 3 is an even function. Evaluate f(-3) in f(x);

Examples 1) Prove that g(x) = x 2 (x 2 - 4) is an even function. A prove statement requires you to start with the LHS and arrive at the RHS. g(x) is even if g(-x) = g(x) by definition At x = a, g(a) = a 2 (a 2 - 4) At x = -a g(-a) = a 2 [a 2 - 4] g(-a) = g(a) Hence, g(x) = x 2 (x 2 – 4) is an even function. g(-a) = (-a) 2 [(-a) 2 - 4] but, g(a) = a 2 (a 2 - 4) 2) Show that is an even function A show statement requires you to evaluate LHS and the RHS and show they are equal. At x = a, At x = -a, Hence, f(-a) = f(a) Therefore is an even function f(x) is even if f(-x) = f(x) by definition

Odd Functions A function, f(x), is an odd function if f(-x) = –f(x), for all x in its domain. An odd function has the important property point symmetry, 180° rotation about the origin. This allows us to plot the top-side of the x-axis, then to complete the sketch, draw the 180° rotated mirror image on the bottom-side of the x-axis. Y-axis Top-side of y = f(x) y = f(x) Y-axis y = f(x) Bottom-side of y = f(x) 180° rotation

Testing for an Odd Function To test whether a function is even or odd or neither, work with f(-x) and compare it to f(x). Usually, it is also better to use x = a and x = -a as the two values for comparison. Examples Show that f(x) = x 3 - 4x is an odd function. Lets use numerical values x = 4 and x = -4. Evaluate f(4) in f(x); f(x) = x 3 – 4x f(4) = (4) 3 – 4(4) f(4) = 48 andf(x) = x 3 – 4x f(-4) = (-4) 3 – 4(-4) f(-2) = -48 Now, f(-4) = – f(4) Hence, f(x) = x 3 - 4x is an odd function. But, is the same true for x = ±4, ±12,.. To be certain, you should test every value of x! This is not practical. A better method, is to use algebraic values of x, say x = ±a f(x) = x 3 – 4x f(a) = (a) 4 – 4(a) f(a) = a 4 – 4a andf(x) = x 4 - 4x f(-a) = (-a) 3 - 4(-a) f(-a) = -a 3 + 4a Now, f(-a) = –f(a) Hence, f(x) = x 3 - 4x is an odd function. Evaluate f(-4) in f(x); But, –f(a) = –(a 3 – 4a) f(-a) = –(a 3 – 4a)

Examples 1) Prove that f(x) = 5x - x 3 is an odd function. f(x) is an odd function if f(-x) = – f(x) At x = a f(x) = 5x - x 3 f(a) = 5a - a 3 At x = -a f(x) = 5x - x 3 f(-a) = 5(-a) - (-a) 3 = -5a - -a 3 = -5a + a 3 = -(5a – a 3 ) = – f(a) Hence, f(-a) = –f(a), Therefore f(x) = 5x - x 3 is an odd function. 2) Show that is an odd function. f(x) is an odd function if f(-x) = – f(x) At x = a At x = -a = – f(a) Hence, f(-a) = –f(a), Therefore is an odd function.

Example: Show that f(x) = x 2 + x is neither an even or odd function. To show or prove that a function is neither odd or even, you need to show both the following two conditions. I f(x) ≠ f(-x) IIf(-x) ≠ – f(x) At x = a, f(x) = x 2 + x f(a) = a 2 + a At x = -a, f(x) = x 2 + x f(-a) = (-a) 2 + (-a) f(-a) = a 2 – a But, – f(a) = –a 2 – a Now, f(a) ≠ f(-a) and f(-a) ≠ – f(a) Hence, f(x) = x 2 + x is neither an even or odd function.
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