The Central Idea
Suppose
\[ f(x)=2x+3. \]If the input is
\[ x=4, \]then we write
\[ f(4). \]This means:
Evaluate the function \(f\) using \(4\) as the input.
Substitute \(4\) wherever the input variable \(x\) appears:
\[ f(4)=2(4)+3. \]Then
\[ f(4)=11. \]So
\[ \boxed{f(4)=11}. \]What Does \(f(x)\) Mean?
The notation
\[ f(x) \]means “the output of function \(f\) at input \(x\).”
It does not mean
\[ f\cdot x. \]The symbol \(f\) names the function, and the expression inside parentheses names the input.
Function Notation Separates the Rule from the Input
Consider
\[ f(x)=x^2+1. \]The rule is:
\[ x\mapsto x^2+1. \]Different inputs produce different evaluations:
\[ f(2)=5, \] \[ f(5)=26, \] \[ f(-3)=10. \]The function itself stays the same while the input changes.
The Letter \(x\) Is a Placeholder
In
\[ f(x)=3x-7, \]the \(x\) represents the input position.
If the input is \(5\), replace every \(x\) with \(5\):
\[ f(5)=3(5)-7=8. \]If the input is \(a\), replace every \(x\) with \(a\):
\[ f(a)=3a-7. \]Function Names Can Change
Functions need not be named \(f\).
For example,
\[ g(x)=x^2-4, \] \[ h(t)=5t+1, \] \[ A(r)=\pi r^2. \]The letter naming the function is simply a label.
Input Variables Can Change Too
Suppose
\[ d(t)=60t. \]The input variable is \(t\), not \(x\).
Then
\[ d(3)=60(3)=180. \]If \(d\) represents distance in miles and \(t\) represents hours, then
\[ d(3)=180\text{ miles}. \]Evaluation Is Substitution
Function evaluation is a structured form of substitution.
If
\[ f(x)=x^2-3x+2, \]then
\[ f(4) \]means replace every occurrence of \(x\) with \(4\):
\[ f(4)=4^2-3(4)+2. \]Then
\[ f(4)=16-12+2=6. \]Use Parentheses When Substituting
Parentheses help preserve the structure of the original formula.
For
\[ f(x)=x^2-5x, \]evaluate at
\[ x=-2. \]Write
\[ f(-2)=(-2)^2-5(-2). \]Then
\[ f(-2)=4+10=14. \]Parentheses prevent sign errors.
Negative Inputs
Suppose
\[ f(x)=x^2+2x. \]Evaluate
\[ f(-3). \]Substitute:
\[ f(-3)=(-3)^2+2(-3). \]Then
\[ f(-3)=9-6=3. \]So
\[ \boxed{f(-3)=3}. \]Why \((-3)^2\) Matters
The expressions
\[ (-3)^2 \]and
\[ -3^2 \]are not the same.
We have
\[ (-3)^2=9, \]while ordinary order of operations gives
\[ -3^2=-(3^2)=-9. \]When substituting a negative input, grouping it in parentheses preserves the intended input.
Fractional Inputs
Suppose
\[ f(x)=4x-1. \]Evaluate
\[ f\left(\frac32\right). \]Substitute:
\[ f\left(\frac32\right) = 4\left(\frac32\right)-1. \]Then
\[ f\left(\frac32\right) = 6-1 = 5. \]Therefore,
\[ \boxed{f\left(\frac32\right)=5}. \]Decimal Inputs
Let
\[ f(x)=2.5x+4. \]Then
\[ f(1.2)=2.5(1.2)+4. \]So
\[ f(1.2)=3+4=7. \]Zero as an Input
Evaluating a function at zero can reveal useful information.
For
\[ f(x)=3x+8, \]we have
\[ f(0)=3(0)+8=8. \]On a graph, this is the output where the input coordinate is zero.
Check the Domain Before Evaluating
Suppose
\[ f(x)=\frac{1}{x-2}. \]Then
\[ f(5)=\frac{1}{5-2}=\frac13. \]But
\[ f(2) \]is undefined because
\[ 2-2=0. \]Evaluation is allowed only for inputs in the domain.
An Undefined Evaluation Is Not Zero
If
\[ f(x)=\frac{1}{x}, \]then
\[ f(0) \]is undefined.
It is not equal to zero.
“Undefined” means the function assigns no output to that input because the input lies outside the domain.
Square-Root Domain Check
Suppose
\[ f(x)=\sqrt{x-1}. \]Then
\[ f(5)=\sqrt{4}=2. \]But over the real numbers,
\[ f(0)=\sqrt{-1} \]is not a real output.
So \(0\) is not in the real domain of this function.
Evaluating an Algebraic Input
The input does not have to be a number.
If
\[ f(x)=2x+5, \]then
\[ f(a)=2a+5. \]If the input is
\[ a+1, \]then
\[ f(a+1)=2(a+1)+5. \]Simplify:
\[ \boxed{f(a+1)=2a+7}. \]Replace Every Occurrence of the Input Variable
Suppose
\[ f(x)=x^2+3x-1. \]Then
\[ f(t) = t^2+3t-1. \]Every \(x\) is replaced by \(t\).
Evaluating \(f(x+1)\)
Suppose
\[ f(x)=x^2. \]Then
\[ f(x+1) = (x+1)^2. \]Expanding:
\[ f(x+1)=x^2+2x+1. \]The entire expression \(x+1\) is the input.
A Common Substitution Error
If
\[ f(x)=x^2+4 \]and the input is
\[ x+1, \]then
\[ f(x+1) = (x+1)^2+4, \]not
\[ x^2+1+4. \]The full input replaces every occurrence of the original placeholder.
Evaluate a More Complicated Input
Let
\[ f(x)=3x-2. \]Find
\[ f(2a-1). \]Substitute:
\[ f(2a-1) = 3(2a-1)-2. \]Simplify:
\[ f(2a-1) = 6a-3-2 = \boxed{6a-5}. \]Function Values Can Be Expressions
Evaluation does not always produce a single number.
If the input itself contains variables, the output may remain algebraic.
For example,
\[ f(x)=x^2+1 \]gives
\[ f(a)=a^2+1. \]This is still a completed evaluation.
Evaluating from a Table
Suppose a function is represented by:
| \(x\) | \(f(x)\) |
|---|---|
| 1 | 4 |
| 2 | 7 |
| 5 | 11 |
Then
\[ f(2)=7. \]No formula is required to evaluate a listed table input.
Input Versus Output Questions
A question such as
\[ f(3)=? \]gives the input and asks for the output.
A question such as
\[ f(x)=11 \]asks for the input or inputs that produce output \(11\).
These are opposite directions of inquiry.
Finding an Input from an Output
Suppose
\[ f(x)=2x+3. \]Find \(x\) if
\[ f(x)=15. \]Write
\[ 2x+3=15. \]Then
\[ 2x=12, \] \[ \boxed{x=6}. \]An Output Can Have More Than One Input
Let
\[ f(x)=x^2. \]Find the inputs satisfying
\[ f(x)=9. \]Then
\[ x^2=9. \]So
\[ \boxed{x=3\text{ or }x=-3}. \]A function gives one output per input, but one output may come from several inputs.
Evaluation Is Not Solving
When we compute
\[ f(4), \]the input is already known. We substitute and calculate.
When we solve
\[ f(x)=4, \]the output is known and the input is unknown.
These are related but different tasks.
Function Notation with Units
Suppose
\[ C(w)=3.20w, \]where \(w\) is pounds and \(C(w)\) is cost in dollars.
Then
\[ C(5)=3.20(5)=16. \]Interpretation:
\[ \boxed{C(5)=\$16}. \]This means 5 pounds cost \$16.
Interpret the Input and Output
If
\[ d(t)=60t, \]then
\[ d(3)=180 \]should not be read merely as “\(d(3)\) equals 180.”
In context, it means:
After 3 hours, the modeled distance is 180 miles.
Function Notation Can Encode Real Questions
If \(T(h)\) gives temperature \(h\) hours after noon, then
\[ T(4) \]means the temperature four hours after noon.
The expression
\[ T(h)=70 \]asks when the temperature is 70 degrees.
Evaluating Piecewise Functions
Consider
\[ f(x)= \begin{cases} x+2, & x<0,\\ x^2, & x\ge0. \end{cases} \]To evaluate, first determine which condition the input satisfies.
Piecewise Example: Negative Input
Using
\[ f(x)= \begin{cases} x+2, & x<0,\\ x^2, & x\ge0, \end{cases} \]find
\[ f(-3). \]Since
\[ -3<0, \]use the first rule:
\[ f(-3)=-3+2=-1. \]Thus,
\[ \boxed{f(-3)=-1}. \]Piecewise Example: Boundary Input
For the same function, find
\[ f(0). \]Since
\[ 0\ge0, \]use the second rule:
\[ f(0)=0^2=0. \]Therefore,
\[ \boxed{f(0)=0}. \]Boundary Symbols Matter
The difference between
\[ x<0 \]and
\[ x\le0 \]determines which rule handles the boundary input.
Piecewise evaluation requires reading inequalities carefully.
Absolute Value as a Piecewise Function
The absolute-value function can be written
\[ |x|= \begin{cases} x, & x\ge0,\\ -x, & x<0. \end{cases} \]So
\[ |-5|=-(-5)=5. \]This is function evaluation through a piecewise rule.
Evaluating Nested Expressions
Suppose
\[ f(x)=x^2+1 \]and we want
\[ f(2+3). \]The input is the whole expression \(2+3\):
\[ f(2+3)=(2+3)^2+1. \]Then
\[ f(5)=26. \]Function Notation and Composition Preview
If
\[ f(x)=x+1 \]and
\[ g(x)=x^2, \]then
\[ g(f(3)) \]means evaluate the inside function first:
\[ f(3)=4. \]Then use that output as the input to \(g\):
\[ g(4)=16. \]So
\[ g(f(3))=16. \]Composition Is Repeated Evaluation
The notation
\[ g(f(x)) \]means that the output of \(f\) becomes the input of \(g\).
This is a natural extension of ordinary function evaluation.
Function Notation with Several Inputs
A function may accept more than one variable.
For example,
\[ A(l,w)=lw. \]To evaluate at
\[ l=5,\qquad w=8, \]write
\[ A(5,8)=5(8)=40. \]Ordered Inputs Matter
For
\[ f(x,y)=2x-y, \]we have
\[ f(3,5)=2(3)-5=1. \]But
\[ f(5,3)=2(5)-3=7. \]The ordered pair \((x,y)\) has a specific input order.
Three-Variable Evaluation
Suppose
\[ V(l,w,h)=lwh. \]Then
\[ V(2,3,5)=2(3)(5)=30. \]The same evaluation principle works for any finite number of input variables.
Evaluating a Formula with Parameters
Suppose
\[ f(x)=ax+b, \]where \(a\) and \(b\) are parameters.
Then
\[ f(3)=3a+b. \]Evaluation substitutes the input variable \(x\), while the parameters remain symbolic unless their values are also supplied.
Distinguish Variables by Role
In
\[ f(x)=ax+b, \]the symbol \(x\) is the function input.
The symbols \(a\) and \(b\) may control which member of a family of functions is being used.
Not every symbol in a formula is necessarily an input variable.
Evaluating a Sequence as a Function
A sequence can be viewed as a function whose domain is usually a set of integers.
Suppose
\[ a(n)=2n+1, \qquad n\in\mathbb{Z}_{>0}. \]Then
\[ a(4)=2(4)+1=9. \]The function notation works the same way even though the domain is discrete.
Evaluation from a Graph
If a graph represents \(y=f(x)\), then \(f(a)\) is the vertical output corresponding to horizontal input \(x=a\).
Graphically, function evaluation means:
- locate the input on the horizontal axis;
- move to the graph;
- read the corresponding output value.
Solving \(f(x)=b\) from a Graph
The reverse question asks which horizontal inputs have output \(b\).
Graphically:
- locate \(b\) on the vertical axis;
- look horizontally for intersections with the graph;
- read the corresponding input values.
There may be zero, one, or several such inputs.
Difference Quotient Preview
Function notation lets us compare outputs at nearby inputs.
For example,
\[ f(x+h)-f(x) \]measures a change in output.
Dividing by the input change \(h\) gives
\[ \frac{f(x+h)-f(x)}{h}, \qquad h\ne0. \]This expression later becomes central to average rates of change and calculus.
Evaluating \(f(x+h)\)
Suppose
\[ f(x)=x^2. \]Then
\[ f(x+h)=(x+h)^2. \]Expand:
\[ f(x+h)=x^2+2xh+h^2. \]This is ordinary substitution with a more complicated input.
Building a Difference Quotient
For
\[ f(x)=x^2, \]we have
\[ f(x+h)-f(x) = (x+h)^2-x^2. \]Simplify:
\[ 2xh+h^2. \]Then
\[ \frac{f(x+h)-f(x)}{h} = \frac{2xh+h^2}{h} = 2x+h, \qquad h\ne0. \]The function notation keeps track of where each input is being evaluated.
Evaluating Does Not Mean Multiplying by the Function Name
If
\[ f(x)=x+3, \]then
\[ f(2)=5. \]It does not mean
\[ f\cdot2. \]The function name \(f\) is not ordinarily treated as a numerical coefficient in this notation.
Do Not Replace Only Some Occurrences
Suppose
\[ f(x)=x^2+4x-1. \]To evaluate \(f(3)\), every occurrence of \(x\) must become \(3\):
\[ f(3)=3^2+4(3)-1. \]Partial substitution changes the rule and produces an incorrect result.
Do Not Confuse \(f(x)^2\) with \(f(x^2)\)
These expressions are generally different.
If
\[ f(x)=x+1, \]then
\[ [f(x)]^2=(x+1)^2, \]while
\[ f(x^2)=x^2+1. \]The location of the exponent matters.
Do Not Confuse \(f(-x)\) with \(-f(x)\)
For
\[ f(x)=x^2+x, \]we have
\[ f(-x)=(-x)^2+(-x)=x^2-x. \]But
\[ -f(x)=-(x^2+x)=-x^2-x. \]These expressions are not generally equal.
Do Not Assume \(f(a+b)=f(a)+f(b)\)
This property does not hold for arbitrary functions.
For
\[ f(x)=x^2, \]we have
\[ f(a+b)=(a+b)^2, \]while
\[ f(a)+f(b)=a^2+b^2. \]These differ by the term
\[ 2ab. \]Special functions may have additive properties, but they must be established rather than assumed.
Worked Example: Polynomial Evaluation
Let
\[ f(x)=2x^3-x^2+4. \]Find
\[ f(-2). \]Substitute:
\[ f(-2) = 2(-2)^3-(-2)^2+4. \]Then
\[ f(-2) = 2(-8)-4+4 = -16. \]Therefore,
\[ \boxed{f(-2)=-16}. \]Worked Example: Rational Function
Let
\[ f(x)=\frac{x+1}{x-3}. \]Find
\[ f(5). \]Substitute:
\[ f(5) = \frac{5+1}{5-3} = \frac62 = 3. \]So
\[ \boxed{f(5)=3}. \]Worked Example: Algebraic Input
Let
\[ f(x)=x^2-2x. \]Find
\[ f(a+1). \]Substitute:
\[ f(a+1) = (a+1)^2-2(a+1). \]Expand:
\[ a^2+2a+1-2a-2. \]Simplify:
\[ \boxed{f(a+1)=a^2-1}. \]Worked Example: Piecewise Evaluation
Let
\[ f(x)= \begin{cases} 2x+1, & x<3,\\ x^2, & x\ge3. \end{cases} \]Find \(f(2)\):
\[ f(2)=2(2)+1=5. \]Find \(f(3)\):
\[ f(3)=3^2=9. \]Thus,
\[ \boxed{f(2)=5,\qquad f(3)=9}. \]Worked Example: Multivariable Function
Let
\[ f(x,y)=x^2+2y. \]Evaluate
\[ f(3,-2). \]Substitute:
\[ f(3,-2)=3^2+2(-2). \]Then
\[ f(3,-2)=9-4=5. \]Therefore,
\[ \boxed{f(3,-2)=5}. \]Worked Example: Context
Suppose
\[ P(t)=1200(1.05)^t \]models a quantity \(t\) years after an initial time.
Then
\[ P(0)=1200. \]This tells us the initial modeled value.
Also,
\[ P(1)=1200(1.05)=1260. \]The notation lets us ask about the model at specific times.
A Reliable Evaluation Workflow
- Identify the function rule.
- Identify the requested input.
- Check that the input belongs to the domain.
- Replace every occurrence of the input variable with the entire input expression.
- Use parentheses around negative, fractional, or algebraic inputs.
- Simplify using ordinary algebra and order of operations.
- Interpret the output and units if the function represents a real situation.
A Reliable Piecewise Workflow
- Locate the input.
- Determine which condition contains that input.
- Use only the corresponding formula.
- Substitute and simplify.
- Check boundary symbols such as \(<\), \(\le\), \(>\), and \(\ge\).
Common Errors
Reading \(f(x)\) as Multiplication
It means function evaluation, not \(f\cdot x\).
Ignoring the Domain
Some requested inputs may be undefined.
Dropping Parentheses Around Negative Inputs
This can reverse signs when powers are involved.
Replacing Only One Occurrence of the Variable
Every occurrence of the input placeholder must be replaced.
Substituting Only Part of an Algebraic Input
The whole expression inside \(f(\,\cdot\,)\) is the input.
Using the Wrong Piecewise Branch
Check the input condition before evaluating.
Assuming \(f(a+b)=f(a)+f(b)\)
This is not a general function law.
Confusing \(f(x)^2\), \(f(x^2)\), and \(f(f(x))\)
These are different constructions and must be read according to their notation.
Practice
1. Basic Evaluation
Let
\[ f(x)=3x+2. \]Find
\[ f(5). \]Check Answer
\[ f(5)=3(5)+2=17. \]Answer:
\[ \boxed{17}. \]2. Negative Input
Let
\[ f(x)=x^2-4x. \]Find
\[ f(-2). \]Check Answer
\[ f(-2)=(-2)^2-4(-2)=4+8=12. \]Answer:
\[ \boxed{12}. \]3. Fractional Input
Let
\[ f(x)=6x-1. \]Find
\[ f\left(\frac12\right). \]Check Answer
\[ f\left(\frac12\right) = 6\left(\frac12\right)-1 = 3-1 = 2. \]Answer:
\[ \boxed{2}. \]4. Algebraic Input
Let
\[ f(x)=2x+3. \]Find
\[ f(a-2). \]Check Answer
\[ f(a-2)=2(a-2)+3. \] \[ =2a-4+3. \] \[ \boxed{2a-1}. \]5. Domain Check
Let
\[ f(x)=\frac{4}{x+1}. \]Evaluate \(f(-1)\).
Check Answer
The denominator becomes zero, so
\[ \boxed{f(-1)\text{ is undefined}}. \]6. Find an Input
Let
\[ f(x)=5x-4. \]Find \(x\) if
\[ f(x)=21. \]Check Answer
\[ 5x-4=21. \] \[ 5x=25. \] \[ \boxed{x=5}. \]7. Piecewise Function
Let
\[ f(x)= \begin{cases} x-1, & x<2,\\ 2x, & x\ge2. \end{cases} \]Find
\[ f(2). \]Check Answer
Since \(2\ge2\), use the second rule:
\[ f(2)=2(2)=4. \]Answer:
\[ \boxed{4}. \]8. Multivariable Function
Let
\[ g(x,y)=3x+y. \]Find
\[ g(2,-5). \]Check Answer
\[ g(2,-5)=3(2)+(-5)=1. \]Answer:
\[ \boxed{1}. \]9. Evaluate \(f(x+h)\)
Let
\[ f(x)=x^2+1. \]Find
\[ f(x+h). \]Check Answer
\[ f(x+h) = (x+h)^2+1. \]Expanded:
\[ \boxed{x^2+2xh+h^2+1}. \]Key Ideas
- \(f(x)\) means the output of function \(f\) at input \(x\).
- Function notation is not multiplication notation.
- Evaluation is systematic substitution.
- The entire expression inside the parentheses is the input.
- Parentheses are especially important for negative, fractional, and algebraic inputs.
- Every occurrence of the input variable must be replaced.
- Inputs must belong to the function's domain.
- Evaluation may produce a number or an algebraic expression.
- Tables, graphs, formulas, and piecewise rules can all be evaluated.
- In piecewise functions, first determine which condition contains the input.
- Finding \(f(a)\) and solving \(f(x)=a\) are opposite-direction questions.
- One output may correspond to several inputs even though each input has only one output.
- Multivariable functions use ordered tuples of inputs.
- Expressions such as \(f(x+h)\) prepare the way for rates of change and calculus.
Where This Leads
Function notation gives us a compact language for asking increasingly sophisticated questions about relationships.
From basic evaluation, the same notation develops into function transformations, composition, inverse functions, solving equations graphically and algebraically, sequences, average rates of change, difference quotients, and eventually limits and derivatives.