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Functions II

Functions II. Odd and Even Functions By Mr Porter. Y-axis. Y-axis. y = f ( x ). y = f ( x ). Right-hand side of f(x). Y-axis is an axis of symmetry. Even Functions. A function, f(x), is an even function if f(x) = f(-x) , for all x in its domain.

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Functions II

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  1. Functions II Odd and Even Functions By Mr Porter

  2. Y-axis Y-axis y = f(x) y = f(x) Right-hand side of f(x) Y-axis is an axis of symmetry 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.

  3. 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 A better method, is to use algebraic values of x, say x = ±a Show that f(x) = x2 - 3 is an even function. Lets use numerical values x = 3 and x = -3. f(x) = x2 - 3 and f(x) = x2 - 3 Evaluate f(3) in f(x); Evaluate f(-3) in f(x); f(a) = (a)2 - 3 f(-a) = (-a)2 - 3 f(x) = x2 - 3 and f(x) = x2 - 3 f(a) = a2 - 3 f(-a) = a2 - 3 f(3) = (3)2 - 3 f(-3) = (-3)2 - 3 Now, f(-a) = f(a) Hence, f(x) = x2 - 3 is an even function. f(3) = 6 f(-3) = 6 Now, f(-3) = f(3) Hence, f(x) = x2 - 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.

  4. 2) Show that is an even function Therefore is an even function Examples 1) Prove that g(x) = x2(x2 - 4) is an even function. A show statement requires you to evaluate LHS and the RHS and show they are equal. A prove statement requires you to start with the LHS and arrive at the RHS. f(x) is even if f(-x) = f(x) by definition g(x) is even if g(-x) = g(x) by definition At x = a, g(a) = a2(a2 - 4) At x = a, At x = -a At x = -a, g(-a) = (-a)2[(-a)2 - 4] g(-a) = a2[a2 - 4] but, g(a) = a2(a2 - 4) g(-a) = g(a) Hence, f(-a) = f(a) Hence, g(x) = x2(x2 – 4) is an even function.

  5. Y-axis Y-axis y = f(x) y = f(x) Top-side of y = f(x) 180° rotation Bottom-side of y = f(x) 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.

  6. 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 A better method, is to use algebraic values of x, say x = ±a Show that f(x) = x3 - 4x is an odd function. Lets use numerical values x = 4 and x = -4. f(x) = x3 – 4x and f(x) = x4 - 4x Evaluate f(4) in f(x); Evaluate f(-4) in f(x); f(a) = (a)4 – 4(a) f(-a) = (-a)3 - 4(-a) f(x) = x3 – 4x and f(x) = x3 – 4x f(a) = a4 – 4a f(-a) = -a3 + 4a f(4) = (4)3 – 4(4) f(-4) = (-4)3 – 4(-4) f(-a) = –(a3 – 4a) f(4) = 48 f(-2) = -48 But, –f(a) = –(a3 – 4a) Now, f(-4) = – f(4) Hence, f(x) = x3 - 4x is an odd function. Now, f(-a) = –f(a) Hence, f(x) = x3 - 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.

  7. Hence, f(-a) = –f(a), Therefore is an odd function. Examples 2) Show that is an odd function. 1) Prove that f(x) = 5x - x3 is an odd function. f(x) is an odd function if f(-x) = – f(x) f(x) is an odd function if f(-x) = – f(x) At x = a f(x) = 5x - x3 At x = a f(a) = 5a - a3 At x = -a f(x) = 5x - x3 At x = -a f(-a) = 5(-a) - (-a)3 = -5a - -a3 = -5a + a3 = -(5a – a3) = – f(a) = – f(a) Hence, f(-a) = –f(a), Therefore f(x) = 5x - x3 is an odd function.

  8. Example: Show that f(x) = x2 + 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) II f(-x) ≠ – f(x) f(x) = x2 + x At x = a, f(a) = a2 + a f(x) = x2 + x At x = -a, f(-a) = (-a)2 + (-a) f(-a) = a2 – a But, – f(a) = –a2 – a Now, f(a) ≠ f(-a) and f(-a) ≠ – f(a) Hence, f(x) = x2 + x is neither an even or odd function.

  9. (1) (2) (3) (4) (5) Exercise Show whether the following functions are EVEN, ODD or NEITHRER. Answer: neither Answer: even Answer: odd Answer: even Answer: odd

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