The ordered pair definition of a function is not all that useful. A function as a formula is a better way to think of a function. Regarding a function as a process is an improvement over thinking of a function as a formula.
In the diagram above, the input x is processed by the function f to produce an output f(x). The input x is known as the argument of f. Replacing x by a specific value, say 3, and running this input through the process to produce an output of f(3) is known as evaluating the function at the point x = 3.
As you have seen, the process can be described by a formula: \[ \begin{align} f(x) &= x^2 + x + 1 \\ \end{align} \]
It can also be described by a rule.
Example The absolute value function can be described as:
The process view of a function is useful for calculations with functions.
Describe the function \(A(r) = \pi r^2\) in words. Plot a few points.
Functions can be added, subtracted and multiplied like ordinary numbers.
Example If \[ f(x) = x^2 \quad \text{and} \quad g(x) = \frac{1}{x+3} \ \] then \[ \begin{align} f(x) + g(x) &= x^2 + \frac{1}{x+3} \\ \end{align} \ \] and the domain is the largest domain common to both functions. Similarly, \[ \begin{align} f(x) \times g(x) &= x^2 \times \frac{1}{x+3} \quad = \frac{x^2}{x+3} \\ \end{align} \ \]
Functions can also be divided, provided the divisor is not 0.
Example The function \[ \begin{align} x \div (x^2-1) &= \frac{x}{x^2-1} \quad \ \\ \end{align} \ \] is defined everywhere except at the points x = ± 1.
In the notation for a function, f(x), the x is just a place holder and could be replaced by any symbol, for example f(t), or even a blank space, f( ). So the function that squares its input can be written as \[ f()=()^2 \] The output of one function can be used as the input of another function.
Example If \[ f(x) = x^2 + x + 1 \quad \ \] it can be rewritten in another form: \[ f() = ()^2 + () + 1 \ \] and another function can be substituted in the ().
Using the output of one function as the input to another is called composing two functions. That is, a function can be a function of another function. If f is a function of g then you write \[ f(g(x)) \] Continuing the example above, if \( g(x) = x+2 \), \[ \begin{align} f(g(x)) &= (g(x))^2 + (g(x)) + 1 \quad \\ &= (x+2)^2 + (x+2) + 1 \quad \\ &= x^2 + 5x + 7 \quad \\ \end{align} \ \]
If \(f(x) = x^2+1\) and \(g(x) = 2x+1\), write down an expression for f(g(x)).
Note that the domain of the composed function depends on the domain of the 'inside' function.
Example If \[ f(x) = \frac{1}{x^2 - 1}\quad \text{and} \quad g(x) = \sqrt{x-2} \ \] then the domain of g(x) is \( x-2 \ge 0 \quad \text{or} \quad x \ge 2 \) and \[ \begin{align} f(g(x)) &= \frac{1}{(\sqrt{x-2})^2 - 1} \\ &= \frac{1}{x-3} \\ \end{align} \ \] so as well as excluding x = 3 from the domain, x must also be greater than or equal to 2.
Guided Examples
O E Qs
Finding the points where a function is zero provides valuable information about the function. Finding those points where \[ \begin{align} f(x) &= 0 \\ \end{align} \ \] means finding the points where f intercepts the x-axis.
If a < b and f(a) < 0 and f(b) > 0 what can you say about the value(s) of f in the interval a < x < b?
Example If you know that the function f is zero at the points 1 and 5, then you know that f must have a maximum or minimum between 1 and 5.
As an extension of this, if you want to find the points where f is equal to some constant, say k, then all you need to do is solve the equation \[ \begin{align} f(x) - k &= 0 \\ \end{align} \ \]
Example to find the points where \[ \ \begin{align} -x^2 + 2x + 3 &= 3 \quad \\ \end{align} \]
Subtracting 3 from both sides: \[ \begin{align} -x^2 + 2x &= 0 \\ -x(x-2) &= 0 \end{align} \ \] so x = 0 or x = 2.
If a function is one to one on its domain, it can be reflected in the line y = x to produce another function, the inverse.
The domain and range of an inverse function are the range and domain of the original function.
To find the equation of an inverse function, substitute x for y and y for x. Then solve for y.
Example To find the inverse of \[ y = 2x + 1 \ \] Interchanging x and y, \[ \begin{align} x &= 2y + 1 \\ 2y &= x - 1 \\ y &= \frac{x-1}{2} \quad \\ \end{align} \ \]
The inverse function is written as f−1(x).
For \(y = x^2, \ x \ge 0\), find the domain and range of the inverse function.
Restricting The Domain
Note that even though some inverse relations may not be functions, they can be made functions by restricting the domain of the original relation so that it is one to one on the restricted domain.
Example The relation \[ y = x^2\ \] is not one to one on its domain so the inverse will not be a function. However, by restricting the domain to x > 0 the inverse relation will be the function \[ y = \sqrt x \ \]