Function Notation in Math
Last Updated :
10 Feb, 2025
Function notation is a precise and simplified way to express the relationship between inputs and outputs. Instead of using the typical y = format, function notation replaces y with a function name, such as f(x), where f represents the function's name, and x is the input variable. This format helps manage multiple functions and better understand how changes in input affect the output.
For example, in the image above the function f(x) = 35x2 + 2, f(x) tells us the output when x is substituted into the equation.
A function is a rule that links two sets of values: the domain (input values) and the range (output values). For each value in the domain, there is exactly one corresponding value in the range.
Mathematically, a function f from set A to set B can be written as:
f : A → B
This means that for every element in the domain A, there is a unique corresponding element in the range B.
Writing Functions Using Notations
Writing functions in notation is a brilliant way of describing functions in mathematics since it is brief. Instead of writing y = 2x + 3, we write f(x) = 2x + 3 Here, f represents an operation that is applied on the quantity x. This is more flexible and better for function manipulation.
In function notation:
- f(x) is a notation that is always stated as “f of x” and it stands for the value of the function at x.
- The letter f also stands for the name of the function while x is the argument of this function.
- The expression inside the parentheses defines the arguments of the function that are taken as the input.
For example, if f(x) = 3x + 5, then:
f(2) = 3(2) + 5 = 6 + 5 = 11
This shows that when x = 2, the output of the function f(x) is 11.
Examples of Function Notation
Function notation can be applied in various contexts:
Linear Function: For f(x) = 4x − 7,
- f(3) = 4(3) − 7 = 12 − 7 = 5.
Quadratic Function: For g(x) = x2 − 4x + 6,
- g(2) = 22 − 4(2) + 6 = 4 − 8 + 6 = 2.
Piecewise Function: h(x) =\begin{cases} x + 2 & \text{if } x \geq 0 \\-x + 2 & \text{if } x < 0\end{cases}
- h(3) = 3 + 2 = 5
- h(-2) = -(-2) + 2 = 4
Common Types of Functions Expressed in Function Notation
Some common types of functions expressed in function notation:
- Linear Functions:
- Form: f(x) = mx +b
- Example: f(x) = 2x + 3
- Quadratic Functions:
- Form: f(x) = ax2 + bx + c
- Example: f(x) = x2 − 4x + 4
- Cubic Functions:
- Form: f(x) = ax3 + bx2 + cx + d
- Example: f(x) = 2x3 − 3x2 + x − 5
- Exponential Functions:
- Form: f(x) = a⋅bx
- Example: f(x) = 3 ⋅ 2x
- Logarithmic Functions:
- Form: f(x) = a ⋅ logb(x) + c
- Example: f(x) = 2 ⋅ log2(x) − 1
- Trigonometric Functions:
- f(x) = sin(x)
- f(x) = cos(x)
- f(x) = tan(x)
- Rational Functions:
- Form: f(x) = p(x)/q(x), where p(x) and q(x) are polynomials.
- Example: f(x) = (x2 − 1)/(x + 1)
- Piecewise Functions:
- Form: f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ x + 1 & \text{if } x \geq 0 \end{cases}
- Absolute Value Functions:
- Form: f(x) = ∣x∣
- Example: f(x) = ∣x − 3∣
Solved Questions on Functions Notations
Question 1: Consider the linear function f(x) = 3x − 4. Find f(2) and determine the x-intercept.
Solution:
To find f(2): f(2) = 3(2) − 4 = 6 − 4 = 2
So, f(2) = 2.
To find the x-intercept, set f(x) = 0 and solve for x:
0 = 3x − 4
3x = 4⟹x = 4/3
Thus, the x-intercept is x = 4/3.
Question 2: Given the quadratic function g(x) = 2x2 − 3x + 1, find the value of g( − 1) and the vertex of the parabola.
Solution:
To find g( − 1): g( − 1) = 2( − 1)2 − 3( − 1) + 1 = 2(1) + 3 + 1 = 6
So, g( − 1) = 6.
The vertex of a quadratic function ax2 + bx + c is given by:
Substituting x = 3/4 into g(x):
g\left(\frac{3}{4}\right) = 2\left(\frac{3}{4}\right)^2 - 3\left(\frac{3}{4}\right) + 1 = \frac{18}{16} - \frac{36}{16} + \frac{16}{16} = \frac{-2}{16} = -\frac{1}{8}
The vertex is (3/4, − 1/8).
Question 3: Find the domain of the rational function h(x) = 5/x2 − 9.
Solution:
The domain of a rational function is all real numbers except where the denominator is zero:
x2 − 9 = 0
(x − 3)(x + 3) = 0
So, x = 3 and x = − 3 make the denominator zero. Thus, the domain is:
x ∈ R ,x = 3,x = − 3
The domain is ( − ∞, − 3)∪( − 3, 3)∪(3, ∞).
Question 4: Evaluate the exponential function f(x) = 2⋅3x at x = − 2 and determine if the function represents growth or decay.
Solution:
To find f( − 2):
f( − 2) = 2⋅3 − 2 = 2⋅ 1/9 = 2/9
So, f( − 2) = 2/9.
Since the base b = 3 is greater than 1, the function represents exponential growth.
Practice Questions of Function Notation
Question 1: Linear Function: Given f(x) = 4x + 5, find f( − 3) and determine the y-intercept.
Question 2: Quadratic Function: If g(x) = x2 − 4x + 4, find the roots of the equation by factoring.
Question 3: Polynomial Function: For the polynomial p(x) = x3 − 6x2 + 11x − 6, find the value of p(1).
Question 4: Exponential Function: Solve for x in the equation 5⋅2𝑥 = 40.
Question 5: Logarithmic Function: If f(x) = log2(x − 1), find the domain of f(x).
Question 6: Piecewise Function: For the function h(x) defined as: h(x) =\begin{cases} 2x + 1 & \text{if } x \geq 0 \\-x^2 + 3 & \text{if } x < 0\end{cases}
Find h(−3) and h(2).
Question 7: Rational Function: Determine the vertical asymptotes of the function h(x) = \frac{x^2 + 1}{x^2 - 4}
Answer Key
- f( − 3) = − 7, y-intercept is 5.
- Roots are x = 2 (double root).
- p(1) = 0.
- x = 3.
- Domain is x>1.
- h( − 3) = − 6, h(2) = 5.
- Vertical asymptotes at x = 2 and x = − 2.
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