Evaluating Functions: Substitution, Simplification, and Interpreting Results

What It Means to Evaluate a Function

To evaluate a function at an input value means: replace the input variable in the formula with that value, then simplify to get a single output. For example, evaluating f(3) means “use input 3,” and the result is the output paired with 3.

A Reliable Routine (Works for Most Functions)

• Step 1: Identify the input (the number or expression inside the parentheses).
• Step 2: Substitute that input everywhere the variable appears.
• Step 3: Use parentheses around the substituted input if it is negative or an expression.
• Step 4: Simplify carefully using correct order of operations.
• Step 5: Check restrictions (especially for rational expressions and radicals).

1) Step-by-Step Substitution Routines

Evaluating Polynomials

Polynomials evaluate by direct substitution and arithmetic. Parentheses matter most when the input is negative or an expression.

Example 1: Simple substitution

Let f(x) = 2x^2 - 3x + 5. Evaluate f(4).

f(4) = 2(4)^2 - 3(4) + 5 = 2(16) - 12 + 5 = 32 - 12 + 5 = 25

Example 2: Negative input (parentheses required)

Let f(x) = x^3 - 4x. Evaluate f(-2).

f(-2) = (-2)^3 - 4(-2) = -8 + 8 = 0

Error Analysis: Polynomial Sign Mistake

Common mistake: writing -2^3 instead of (-2)^3.

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• -2^3 means -(2^3) = -8.
• (-2)^3 means -8 as well here, but for even powers it changes: (-2)^2 = 4 while -2^2 = -(4) = -4.

Evaluating Rational Expressions

For rational expressions, substitute first, then simplify, but also check that you are not dividing by zero.

Example 3: Substitute and simplify

Let g(x) = (x^2 - 9)/(x - 3). Evaluate g(5).

g(5) = (5^2 - 9)/(5 - 3) = (25 - 9)/2 = 16/2 = 8

Example 4: Input that breaks the function

Evaluate g(3) for g(x) = (x^2 - 9)/(x - 3).

g(3) = (3^2 - 9)/(3 - 3) = (9 - 9)/0 = 0/0 (undefined)

Even though the numerator becomes 0, the denominator becomes 0 too, so the expression is not defined at x = 3. The correct result is: undefined.

Error Analysis: Canceling Too Early

Common mistake: factoring and canceling, then plugging in the restricted value.

(x^2 - 9)/(x - 3) = (x - 3)(x + 3)/(x - 3) = x + 3  (only for x ≠ 3)

If you cancel first and then compute 3 + 3 = 6, you would incorrectly claim g(3)=6. The cancellation changes the expression’s form but does not remove the original restriction x ≠ 3.

Evaluating Radicals

With radicals, substitute and simplify, but also ensure the radicand (the expression under the radical) meets any domain requirement (for real-number outputs, square roots require the radicand to be nonnegative).

Example 5: Square root evaluation

Let h(x) = sqrt(x + 1). Evaluate h(8).

h(8) = sqrt(8 + 1) = sqrt(9) = 3

Example 6: Input that makes the radicand negative

Evaluate h(-3) for h(x) = sqrt(x + 1).

h(-3) = sqrt(-3 + 1) = sqrt(-2)

Over the real numbers, sqrt(-2) is not real, so the function has no real output at x = -3.

Error Analysis: Dropping Parentheses Under a Radical

Common mistake: if h(x) = sqrt(x^2 - 4x), evaluating at x = -1 as sqrt(-1^2 - 4(-1)) without parentheses can cause confusion. Write it as:

h(-1) = sqrt((−1)^2 − 4(−1)) = sqrt(1 + 4) = sqrt(5)

2) Parentheses to Avoid Sign Errors (Especially f(-x) and f(-2))

Evaluating f(-2): Treat -2 as the Input Everywhere

If f(x) = x^2 + 6x + 1, then:

f(-2) = (-2)^2 + 6(-2) + 1 = 4 - 12 + 1 = -7

Error Analysis: The “Only Substitute Once” Mistake

Common mistake: substituting into only one term, such as writing f(-2) = (-2)^2 + 6x + 1. Substitution must replace every occurrence of the variable.

Evaluating f(-x): Replace x with (-x) as an Expression

f(-x) is not a number; it is a new expression formed by substituting -x for x. This is a common place for sign errors.

Example 7: f(-x) with even and odd powers

Let f(x) = x^3 - 2x^2 + 5x. Find f(-x).

f(-x) = (-x)^3 - 2(-x)^2 + 5(-x)

= -x^3 - 2x^2 - 5x

Notice the pattern: odd powers change sign, even powers do not, but you should still show parentheses during substitution to avoid mistakes.

Example 8: f(-x) inside a rational expression

Let p(x) = (x - 1)/(x + 2). Find p(-x).

p(-x) = ((-x) - 1)/((-x) + 2) = (-(x + 1))/(2 - x)

You may rewrite it in different but equivalent ways, for example multiplying numerator and denominator by -1:

(-(x + 1))/(2 - x) = (x + 1)/(x - 2)

Error Analysis: Losing a Negative in the Denominator

Common mistake: writing ((-x)+2) = -(x+2). That is incorrect because:

(-x) + 2 = 2 - x  (not -(x+2))

3) Evaluating Piecewise-Defined Functions (Condition Checks)

For a piecewise-defined function, you must choose the correct formula based on the input. Always check which condition the input satisfies before substituting.

Routine for Piecewise Evaluation

• Step 1: Identify the input value.
• Step 2: Compare it to each condition (like x < 0, x ≥ 2, etc.).
• Step 3: Select the matching rule.
• Step 4: Substitute and simplify.

Example 9: Basic piecewise evaluation

Let

f(x) = { 2x + 1,  x < 0  ;  x^2,  x ≥ 0 }

Evaluate f(-3) and f(2).

For x = -3:  -3 < 0, so use 2x + 1:  f(-3) = 2(-3) + 1 = -6 + 1 = -5

For x = 2:  2 ≥ 0, so use x^2:  f(2) = (2)^2 = 4

Example 10: Boundary values matter (≤ vs <)

Let

g(x) = { x + 4,  x ≤ 1  ;  3x - 2,  x > 1 }

Evaluate g(1) and g(1.5).

g(1): since 1 ≤ 1, use x + 4 → g(1) = 1 + 4 = 5

g(1.5): since 1.5 > 1, use 3x - 2 → g(1.5) = 3(1.5) - 2 = 4.5 - 2 = 2.5

Error Analysis: Choosing the Wrong Piece

Common mistake: substituting first and checking the condition later. The condition refers to the input itself, not the output. For example, if the condition is x ≥ 0, you decide using the input value, not by seeing whether the computed output is nonnegative.

4) Interpreting the Meaning of an Output (Context-Like Statements)

After evaluating, interpret the result as “the output corresponding to that input.” In many applications, the input represents an amount, time, length, or count, and the output represents a related quantity.

Example 11: Output as a computed quantity

Suppose C(x) = 12x + 50 models a total cost in dollars for producing x items, where 50 is a fixed setup cost and 12 is the cost per item. Evaluate C(8):

C(8) = 12(8) + 50 = 96 + 50 = 146

Interpretation: When 8 items are produced, the total cost is $146.

Example 12: Interpreting “undefined” or “no real output”

If R(x) = 100/(x - 5) and you try to evaluate R(5), the denominator becomes 0, so the output is undefined.

Interpretation: The model does not produce a meaningful value at input 5 (it is outside the allowed inputs). In context, that often means the situation described by the formula cannot occur or the model breaks at that input.

Example 13: Interpreting piecewise outputs

If a piecewise function uses different rules for different input ranges, the output tells you which rule applies at that input and what value results. For example, if a pricing rule changes after a threshold, evaluating at an input just below and just above the threshold can show the change in output behavior.

Mixed Practice Sets (Progressive)

Set A: Straightforward Substitution (Polynomials)

• 1) f(x) = 3x - 7. Find f(10).
• 2) f(x) = x^2 - 5x + 6. Find f(2).
• 3) f(x) = 2x^3 + x. Find f(-1).
• 4) f(x) = -4x^2 + 9. Find f(-3).

Set B: Parentheses and f(-x)

• 5) f(x) = x^2 + 2x. Find f(-5).
• 6) f(x) = x^3 - x. Find f(-x) and simplify.
• 7) f(x) = 2x^2 - 3x + 1. Find f(-x) and simplify.
• 8) f(x) = (x + 4)^2. Find f(-2).

Set C: Rational Expressions (Include Restrictions)

• 9) g(x) = (x + 1)/(x - 2). Find g(5).
• 10) g(x) = (x^2 - 16)/(x + 4). Find g(0).
• 11) g(x) = (x^2 - 16)/(x + 4). Find g(-4) (state what happens).
• 12) g(x) = 6/(2x - 3). Find g(1.5) (state what happens).

Set D: Radicals (Check for Real Outputs)

• 13) h(x) = sqrt(x + 9). Find h(7).
• 14) h(x) = sqrt(x + 9). Find h(-12) over the real numbers (state what happens).
• 15) h(x) = sqrt(4x - 1). Find h(3).
• 16) h(x) = sqrt(4x - 1). Find h(0) over the real numbers (state what happens).

Set E: Piecewise Functions (Condition Checks)

• 17) f(x) = { x + 2, x < 1 ; 2x - 1, x ≥ 1 }. Find f(0) and f(1).
• 18) g(x) = { x^2, x ≤ -2 ; 3, -2 < x < 2 ; 2x, x ≥ 2 }. Find g(-2), g(0), and g(2).
• 19) p(x) = { (x+1)/(x-1), x ≠ 1 ; 0, x = 1 }. Find p(1) and p(2).
• 20) q(x) = { sqrt(x), x ≥ 0 ; x + 1, x < 0 }. Find q(-4) and q(9).

Set F: Diagnose the Mistake (Error Analysis Practice)

For each item, identify the error and correct it.

• 21) If f(x)=x^2-6x, a student writes: f(-3) = -3^2 - 6(-3) = -9 + 18 = 9.
• 22) If g(x)=(x^2-1)/(x-1), a student writes: g(1)=2 because (x^2-1)/(x-1) = x+1.
• 23) If h(x)=sqrt(x-5), a student writes: h(2)=sqrt(3).
• 24) If p(x)={2x, x<0; x+2, x≥0}, a student evaluates p(-1) using x+2.