Evaluating Functions

Core Principles of Evaluating Functions

• Function Notation and Definitions:

  • A function represents a unique mathematical relation where each input value from the domain corresponds to exactly one output value in the range.

  • The notation $f(x)$ is read as "$f$ of $x$" and denotes the output value of the function $f$ evaluated at the input variable $x$.

  • In function notation, $x$ serves as the independent variable (input), while $f(x)$ represents the dependent variable (output), conventionally plotted along the $y$-axis on the Cartesian coordinate plane.

• Evaluating Functions for a Given Input ($f(a)$):

  • Evaluating $f(a)$ requires direct substitution of the numerical value $a$ into every position where the variable $x$ appears in the defining equation.

  • Following substitution, standard order of operations (PEMDAS: Parentheses, Exponents, Multiplication and Division, Addition and Subtraction) must be applied to compute the final scalar value.

  • Crucial sign rule for exponentiation: $-x^2$ evaluates as $-(x^2)$, whereas $(-x)^2$ evaluates as a positive number.

  • Absolute value operations evaluate as $|u| = u$ when $u \ge 0$, and $|u| = -u$ when $u < 0$.

• Solving for an Unknown Input Given an Output ($f(x) = c$):

  • When given an output equation such as $f(x) = c$, the objective is reversed: replace $f(x)$ with the scalar value $c$ and solve the resulting equation for the independent variable $x$.

  • This requires inverse algebraic operations, such as adding or subtracting constants from both sides, followed by multiplying or dividing by coefficients.

• Evaluating Functions from Graphical Representations:

  • To find $f(a)$ graphically, locate the input value $a$ along the horizontal $x$-axis, move vertically to intersect the curve or line of $f(x)$, and read the corresponding value on the vertical $f(x)$-axis.

  • To solve $f(x) = c$ graphically, locate the output value $c$ on the vertical $f(x)$-axis, move horizontally to intersect the graph, and read the corresponding $x$-coordinate(s) on the horizontal axis.

Direct Evaluation of Algebraic Functions

• Problem A: Quadratic Function Evaluation

  • Given function:     $f(x) = x^2 - 8x - 12$

  • Objective:     Find $f(-1)$.

  • Step-by-step substitution:     $f(-1) = (-1)^2 - 8(-1) - 12$

  • Simplification of exponents and products:     $(-1)^2 = 1$     $-8(-1) = 8$

  • Arithmetic combination:     $f(-1) = 1 + 8 - 12$     $f(-1) = 9 - 12$     $f(-1) = -3$

• Problem B: Linear Function Evaluation

  • Given function:     $f(x) = -4x + 20$

  • Objective:     Find $f(5)$.

  • Step-by-step substitution:     $f(5) = -4(5) + 20$

  • Simplification of product:     $-4(5) = -20$

  • Arithmetic combination:     $f(5) = -20 + 20$     $f(5) = 0$

• Problem C: Linear Function with Fractional Coefficient

  • Given function:     $f(x) = \frac{2}{3}x + 7$

  • Objective:     Find $f(-12)$.

  • Step-by-step substitution:     $f(-12) = \frac{2}{3}(-12) + 7$

  • Simplification of fractional multiplication:     $\frac{2}{3}(-12) = 2(-4) = -8$

  • Arithmetic combination:     $f(-12) = -8 + 7$     $f(-12) = -1$

• Problem D: Absolute Value Function Evaluation

  • Given function:     $f(x) = |9 - 2x|$

  • Objective:     Find $f(7)$.

  • Step-by-step substitution:     $f(7) = |9 - 2(7)|$

  • Multiplication inside absolute value:     $2(7) = 14$

  • Subtraction inside absolute value:     $9 - 14 = -5$

  • Application of absolute value:     $f(7) = |-5|$     $f(7) = 5$

• Problem E: Quadratic Function with Leading Coefficient

  • Given function:     $f(x) = 2x^2 + x - 10$

  • Objective:     Find $f(-4)$.

  • Step-by-step substitution:     $f(-4) = 2(-4)^2 + (-4) - 10$

  • Simplification of exponent:     $(-4)^2 = 16$

  • Simplification of product:     $2(16) = 32$

  • Arithmetic combination:     $f(-4) = 32 - 4 - 10$     $f(-4) = 28 - 10$     $f(-4) = 18$

• Problem F: Cubic Function Evaluation

  • Given function:     $f(x) = x^3 + 4x$

  • Objective:     Find $f(-2)$.

  • Step-by-step substitution:     $f(-2) = (-2)^3 + 4(-2)$

  • Evaluation of powers and products:     $(-2)^3 = -8$     $4(-2) = -8$

  • Arithmetic combination:     $f(-2) = -8 - 8$     $f(-2) = -16$

• Problem G: Linear Function with Fractional Subtraction

  • Given function:     $f(x) = 15 - \frac{5}{2}x$

  • Objective:     Find $f(6)$.

  • Step-by-step substitution:     $f(6) = 15 - \frac{5}{2}(6)$

  • Multiplication of fraction by integer:     $\frac{5}{2}(6) = 5(3) = 15$

  • Arithmetic combination:     $f(6) = 15 - 15$     $f(6) = 0$

• Problem H: Absolute Value of a Quadratic Expression

  • Given function:     $f(x) = |-20 - x^2|$

  • Objective:     Find $f(-8)$.

  • Step-by-step substitution:     $f(-8) = |-20 - (-8)^2|$

  • Evaluation of base exponent:     $(-8)^2 = 64$

  • Subtraction inside absolute value:     $-20 - 64 = -84$

  • Application of absolute value:     $f(-8) = |-84|$     $f(-8) = 84$

• Problem I: Quadratic Function with Leading Negative Coefficient

  • Given function:     $f(x) = -5x^2 + 20$

  • Objective:     Find $f(3)$.

  • Step-by-step substitution:     $f(3) = -5(3)^2 + 20$

  • Exponentiation followed by multiplication:     $3^2 = 9$     $-5(9) = -45$

  • Arithmetic combination:     $f(3) = -45 + 20$     $f(3) = -25$

• Problem J: Negated Leading Quadratic Term

  • Given function:     $f(x) = -x^2 + 5x + 13$

  • Objective:     Find $f(7)$.

  • Step-by-step substitution:     $f(7) = -(7)^2 + 5(7) + 13$

  • Order of operations distinction between $-(7^2)$ and $(-7)^2$:     $-(7)^2 = -49$     $5(7) = 35$

  • Arithmetic combination:     $f(7) = -49 + 35 + 13$     $f(7) = -14 + 13$     $f(7) = -1$

Inverse Algebraic Evaluation: Solving for Unknown Input Values

• Problem K: Linear Equation with Negative Slope

  • Given function:     $f(x) = -3x - 7$

  • Condition:     $f(x) = -61$

  • Objective:     Find $x$.

  • Setting up the linear equation:     $-3x - 7 = -61$

  • Adding $7$ to both sides:     $-3x = -61 + 7$     $-3x = -54$

  • Dividing both sides by $-3$:     $x = \frac{-54}{-3}$     $x = 18$

• Problem L: Linear Equation Involving Variable Subtraction

  • Given function:     $f(x) = 14 - 2x$

  • Condition:     $f(x) = 20$

  • Objective:     Find $x$.

  • Setting up the equation:     $14 - 2x = 20$

  • Subtracting $14$ from both sides:     $-2x = 20 - 14$     $-2x = 6$

  • Dividing both sides by $-2$:     $x = \frac{6}{-2}$     $x = -3$

• Problem M: Rational Linear Expression

  • Given function:     $f(x) = \frac{x - 26}{2}$

  • Condition:     $f(x) = 29$

  • Objective:     Find $x$.

  • Setting up the equation:     $\frac{x - 26}{2} = 29$

  • Multiplying both sides by the denominator $2$:     $x - 26 = 29 \times 2$     $x - 26 = 58$

  • Adding $26$ to both sides:     $x = 58 + 26$     $x = 84$

• Problem N: Negated Variable Equation

  • Given function:     $f(x) = -x - 11$

  • Condition:     $f(x) = 5$

  • Objective:     Find $x$.

  • Setting up the equation:     $-x - 11 = 5$

  • Adding $11$ to both sides:     $-x = 5 + 11$     $-x = 16$

  • Dividing both sides by $-1$:     $x = -16$

• Problem O: Fractional Slope Linear Equation

  • Given function:     $f(x) = \frac{1}{4}x - 8$

  • Condition:     $f(x) = -\frac{7}{2}$

  • Objective:     Find $x$.

  • Setting up the equation:     $\frac{1}{4}x - 8 = -\frac{7}{2}$

  • Adding $8$ to both sides using a common denominator:     $\frac{1}{4}x = -\frac{7}{2} + \frac{16}{2}$     $\frac{1}{4}x = \frac{9}{2}$

  • Multiplying both sides by $4$:     $x = \frac{9}{2} \times 4$     $x = 18$

• Problem P: Two-Step Linear Equation

  • Given function:     $f(x) = 8x + 61$

  • Condition:     $f(x) = -139$

  • Objective:     Find $x$.

  • Setting up the equation:     $8x + 61 = -139$

  • Subtracting $61$ from both sides:     $8x = -139 - 61$     $8x = -200$

  • Dividing both sides by $8$:     $x = \frac{-200}{8}$     $x = -25$

• Problem Q: Two-Step Equation with Constant Subtraction

  • Given function:     $f(x) = 3x - 29$

  • Condition:     $f(x) = -14$

  • Objective:     Find $x$.

  • Setting up the equation:     $3x - 29 = -14$

  • Adding $29$ to both sides:     $3x = -14 + 29$     $3x = 15$

  • Dividing both sides by $3$:     $x = \frac{15}{3}$     $x = 5$

• Problem R: Negative Fractional Coefficient Equation

  • Given function:     $f(x) = -\frac{3}{5}x - 17$

  • Condition:     $f(x) = -2$

  • Objective:     Find $x$.

  • Setting up the equation:     $-\frac{3}{5}x - 17 = -2$

  • Adding $17$ to both sides:     $-\frac{3}{5}x = -2 + 17$     $-\frac{3}{5}x = 15$

  • Multiplying both sides by the reciprocal $-\frac{5}{3}$:     $x = 15 \times \left(-\frac{5}{3}\right)$     $x = -25$

• Problem S: Variable Subtraction with Fractional Rate

  • Given function:     $f(x) = -11 - \frac{1}{6}x$

  • Condition:     $f(x) = -25$

  • Objective:     Find $x$.

  • Setting up the equation:     $-11 - \frac{1}{6}x = -25$

  • Adding $11$ to both sides:     $-\frac{1}{6}x = -25 + 11$     $-\frac{1}{6}x = -14$

  • Multiplying both sides by $-6$:     $x = -14 \times (-6)$     $x = 84$

• Problem T: Linear Equation with Negative Slope and Positive Constant

  • Given function:     $f(x) = -5x + 9$

  • Condition:     $f(x) = 89$

  • Objective:     Find $x$.

  • Setting up the equation:     $-5x + 9 = 89$

  • Subtracting $9$ from both sides:     $-5x = 89 - 9$     $-5x = 80$

  • Dividing both sides by $-5$:     $x = \frac{80}{-5}$     $x = -16$

Graphical Evaluation of Functions

• Problem U: Evaluating a Linear Function from a Graph

• Objective:     Find $f(-4)$.

• Graphical reading procedure:

  • Locate the input value $x = -4$ on the horizontal axis (four grid units to the left of the origin).

  • Follow the vertical grid line down to intersect the graphed line.

  • The intersection occurs exactly at the point $(-4, -3)$.

• Analytical verification:

  • The graphed line passes through the $y$-intercept $(0, -1)$ and the $x$-intercept $(2, 0)$.

  • Slope formula calculation:       $m = \frac{0 - (-1)}{2 - 0} = \frac{1}{2}$

  • Line equation in slope-intercept form:       $f(x) = \frac{1}{2}x - 1$

  • Direct evaluation at $x = -4$:       $f(-4) = \frac{1}{2}(-4) - 1 = -2 - 1 = -3$

• Final value:     $f(-4) = -3$

  • Problem V: Evaluating a Quadratic Function from a Graph

• Objective:     Find $f(2)$.

• Graphical reading procedure:

  • Locate the input value $x = 2$ on the horizontal axis (two grid units to the right of the origin).

  • Move vertically along the grid line $x = 2$ upward until intersecting the curve of the parabola.

  • The intersection corresponds to the vertical level $y = 5$, yielding the coordinate $(2, 5)$.

• Analytical verification:

  • The parabola has its vertex at $(-1, -4)$ and opens upward with standard vertical stretch factor $a = 1$.

  • Vertex form of the quadratic function:       $f(x) = (x + 1)^2 - 4$

  • Intercept checks confirm consistency:       $f(0) = (0 + 1)^2 - 4 = -3$       $f(1) = (1 + 1)^2 - 4 = 0$       $f(-3) = (-3 + 1)^2 - 4 = 0$

  • Evaluating at $x = 2$:       $f(2) = (2 + 1)^2 - 4 = 3^2 - 4 = 9 - 4 = 5$

• Final value:     $f(2) = 5$

  • Problem W: Solving for Input from a Linear Graph

• Objective:     If $f(x) = 4$, find $x$.

• Graphical reading procedure:

  • Identify the output value $f(x) = 4$ on the vertical axis (four grid units above the origin).

  • Trace horizontally along the line $y = 4$ to intersect the graphed line.

  • The intersection occurs at the grid point $(-1, 4)$.

  • Project down to the horizontal axis to find the corresponding input value $x = -1$.

• Analytical verification:

  • The line has a $y$-intercept at $(0, 1)$ and passes through $(1, -2)$.

  • Slope formula calculation:       $m = \frac{-2 - 1}{1 - 0} = -3$

  • Line equation in slope-intercept form:       $f(x) = -3x + 1$

  • Setting the function equal to the specified output:       $-3x + 1 = 4$       $-3x = 3$       $x = -1$

• Final value:     $x = -1$

  • Problem X: Solving for Input from a Polynomial Graph

• Objective:     If $f(x) = -3$, find $x$.

• Graphical reading procedure:

  • Locate the horizontal line corresponding to output $f(x) = -3$ (three grid units below the horizontal axis).

  • Follow this line across to determine where it intersects the graphed cubic polynomial curve.

  • One clear intersection is the local minimum of the curve, which rests tangentially on the vertical axis at $(0, -3)$, providing the input solution $x = 0$.

  • The downward-extending branch on the left also intersects the line $y = -3$ at the grid point $(-3, -3)$, providing the additional solution $x = -3$.

• Analytical verification:

  • The curve represents the cubic function:       $f(x) = x^3 + 3x^2 - 3$

  • First derivative critical point test confirms local extrema:       $f'(x) = 3x^2 + 6x = 3x(x + 2)$       $f'(-2) = 0 \implies f(-2) = (-2)^3 + 3(-2)^2 - 3 = 1$       $f'(0) = 0 \implies f(0) = 0^3 + 3(0)^2 - 3 = -3$

  • Setting the cubic expression equal to $-3$:       $x^3 + 3x^2 - 3 = -3$       $x^3 + 3x^2 = 0$       $x^2(x + 3) = 0$

  • Solutions to the equation:       $x = 0$       $x = -3$

• Final value:     $x = 0$ (or $x = -3$)