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David Terr
Ph.D. Math, UC Berkeley

 

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1.7. Transformations of Functions

There are several useful ways of transforming functions, including reflections, translations, and scale transformations. First we consider reflections. For a given function f(x), consider the function g(x) = -f(x). This is simply the negative of f(x). In other words, for every image y of x under f, -y is the corresponding image under g = -f. As a result, the graph of g is the reflection of f about the x-axis. What about h(x) = f(-x)? If y is the image of x under f, then it is easy to see that y is the corresponding image of -x under h. The graph of h is the reflection of f about the y-axis.

Some functions remain unchanged when reflected about the y-axis. In other words, these functions are symmetric about the y-axis. Such functions are called even functions. Mathematically speaking, a function f(x) is even if f(-x) = f(x) for all x in the domain of f. The reason for this name is that a polynomial f(x) is even if and only if all its nonzero coefficients correspond to even powers of x, i.e. if and only if f(x) has the form f(x) = a2nx2n + a2n-2x2n-2 + ... + a4x4 + a2x2 + a0 for some nonnegative integer n. The absolute value function f(x) = |x| is another example of an even function.

An equally important class of functions are odd functions. These are functions f(x) satisfying f(-x) = -f(x) for all x in the domain of f. Odd functions are symmetric about the origin, i.e. if the point (x,y) lies on the graph of f, then the point (-x, -y) also lies on the graph of f. The reason for this name is that a polynomial f(x) is odd if and only if all its nonzero coefficients correspond to odd powers of x, i.e. if and only if f(x) has the form f(x) = a2n+1x2n+1 + a2n-1x2n-1 + ... + a5x5 + a3x3 + a1x for some nonnegative integer n.

It turns out that every function defined over all real numbers has a unique decomposition into even and odd parts. The decomposition is as follows. Given a function f(x) defined over the reals, let fe(x) = [f(x) + f(-x)] / 2 and fo(x) = [f(x) - f(-x)] / 2. Then it is completely straightforward to verify that fe is even, fo is odd, and that f = fe + fo. We leave this as an exercise.

Given a function f(x) and a real constant a, consider the function g(x) = f(x - a). If y = f(x) is the image of x under f, then y = g(x + a) is the image of x + a under g. In other words, g acts the same way on every input x+a as f acts on x. As a result, the graph of g looks the same as that of f, but shifted by a distance a along the x-axis.

Given a function f(x) and a real constant b, consider the function g(x) = f(x) + b. If y = f(x) is the image of x under f, then y + b = g(x) is the image of x under g. The graph of g looks the same as that of f, but shifted by a distance b along the y-axis.

Given a function f(x) and a positive real constant c, consider the function g(x) = f(x/c). If y = f(x) is the image of x under f, then y = g(cx) is the image of cx under g. In other words, g acts the same way on every input cx as f acts on x. As a result, the graph of g looks the same as that of f, but stretched out by a factor of c along the x-axis. (If c is less than 1, then the graph of g is compressed along the x-axis.)

Given a function f(x) and a positive real constant d, consider the function g(x) = d f(x). If y = f(x) is the image of x under f, then dy = g(x) is the image of x under g. The graph of g looks the same as that of f, but stretched out by a factor of d along the y-axis. (If d is less than 1, then the graph of g is compressed along the y-axis.)

 

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