Question

Odd and Even Functions An even function has the property f(x) =f(-x). Consider the function f(x) Now, f (-a)-(-a)"-d f(a) An odd function has the property f(-x)-f(x). Consider the function f(x) Now, f (-a) = (-a)' =-a3 =-f(a) Declarative & Procedural Knowledge Comment on the meaning of the definitions of even and odd functions in term of transformations. (i) (ii) Show that functions of the formx) are even. bx2 +c Show, that f(x) = asin xis odd and g(x) = b cos xīs even on [-M] (İİİ) Consider functions of the form f(x)-(a-) (iv) Sketch the function for the case when m = 2, n = 1 and a is a positive constant. Nominate a value for your choice of a. Show algebraically that there is always a stationary point on the interval [0, a]. (v)

Answer 1

Even function if a function is reflected about the y axis and remains unchanged Then the function is said to be an even function f(x) is even if f(- x) = f(x) odd function An odd function is that in which theme is a rotation of 180 about the origin any point in the first Quadrant is mapped on to a point in the 3rd Quadrant if f(x) is odd when f(- x) = -f(x) \r\n To show that the function f(x) = a/bx^2 + C is even replace x by - x and simplify f(x) = a/b(- x)^2 + C = a/bx^2 + C = f(x) Because (- x)^2 = x^2 Therefore f(- x) = f(x) The function even f(x) = a sin x (- pi/2, pi/2) restricted Interval f(- x) = a sin (- x) = -a sin x = -f (x) Therefore f(- x) = -f(x) The function is odd g(x) = b cos theta over restricted domain (- pi/2 pi/2) g(x) = b cos theta g(- x) = b cos (- theta) = b cos theta = g (x) Because cos (- theta) = cos theta Therefore g(- x) = g(x) The function g is even Given f(x) = x^m (a - x)^n where m = 2 n = 1 and let a = 1 then f(x) = x^2 (1 - x) = x^2 - x^3 The graph is show separately f(x) = x^2 (a - x)