Function notation and evaluation

Learning goals

Determines whether a relation defines a function.
Uses function notation correctly.
Evaluates a function at given inputs.
Interprets inputs and outputs in context.

Difficulty: 1 – 5

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Function notation and evaluation 1

The linear function is $f(x)=0.4x+1.2$. Find each function value. 1) $f(2)$ 2) $f(-3)$

Hints

- Function notation tells you which input replaces $x$. - Use parentheses when substituting a negative input. - Multiply before adding.

Solution

1. Substitute $x=2$: $f(2)=0.4(2)+1.2=2$. 2. Substitute $x=-3$: $f(-3)=0.4(-3)+1.2=0$.

Answer

1) $f(2)=2$ 2) $f(-3)=0$

Function notation and evaluation 2

A small ball's path is modeled by $h(x) = -0.5x^2 + 2x + 1$, where $x$ is horizontal distance and $h(x)$ is height, both in meters. Evaluate the function to find the ball's height at: 1) $x = 2\,\text{m}$ 2) $x = 4\,\text{m}$

Hints

- Substitute the given input into the function. - Apply the exponent before multiplication and addition. - The negative coefficient is outside the square.

Solution

1. Substitute $x = 2$: $h(2) = -0.5 \cdot 2^2 + 2 \cdot 2 + 1 = -2 + 4 + 1 = 3$. The height is $3\,\text{m}$. 2. Substitute $x = 4$: $h(4) = -0.5 \cdot 4^2 + 2 \cdot 4 + 1 = -8 + 8 + 1 = 1$. The height is $1\,\text{m}$.

Answer

1) $3\,\text{m}$ 2) $1\,\text{m}$

Function notation and evaluation 3

The table defines a function $f$. <table><tr><td>$x$</td><td>$-3$</td><td>$0$</td><td>$2$</td><td>$5$</td></tr><tr><td>$f(x)$</td><td>$7$</td><td>$1$</td><td>$-3$</td><td>$-9$</td></tr></table> a) Find $f(2)$. b) Find $f(-3)$. c) Write the ordered pair represented by $f(5)=-9$.

Hints

- In a function table, match each input in the top row with the output directly below it. - In $f(a)=b$, $a$ is the input and $b$ is the output. - An input-output pair can be written as an ordered pair $(x,y)$.

Solution

1. In the column where $x=2$, the output is $-3$, so $f(2)=-3$. 2. In the column where $x=-3$, the output is $7$, so $f(-3)=7$. 3. The statement $f(5)=-9$ means input $5$ has output $-9$, so the ordered pair is $(5,-9)$.

Answer

a) $-3$ b) $7$ c) $(5,-9)$

Function notation and evaluation 4

Which of these relations is a function? Explain your choice. Relation A: $\{(0,2),(1,3),(0,4)\}$ Relation B: $\{(0,2),(1,3),(2,4)\}$

Hints

- Look at the first number in each ordered pair. - Can one input have two different outputs in a function?

Solution

1. In relation A, input $0$ has two outputs, $2$ and $4$. A function cannot assign two outputs to one input. 2. In relation B, each input appears with exactly one output, so B is a function.

Answer

Relation B is a function. Relation A is not, because input $0$ has two outputs.

Function notation and evaluation 5

Which graph shows $y$ as a function of $x$, a) or b)? Explain why the other graph is not a function.

Hints

- Imagine drawing a vertical line through each graph. - Does any vertical line meet a graph in more than one point?

Solution

1. On graph a), each x-coordinate has one y-coordinate, so it represents a function. 2. On graph b), the points $(0,2)$ and $(0,-2)$ both lie on the circle. One input has two outputs, so it is not a function of $x$.

Answer

Graph a) represents a function. Graph b) does not, because $x=0$ has outputs $2$ and $-2$.

Function notation and evaluation 6

A library charges $\$2$ for each day a book is late, plus a one-time $\$1$ late fee. The total charge is $L(d)=2d+1$, where $d$ is the number of late days. What does $L(3)$ mean? Find its value and include the unit.

Hints

- What quantity goes into the function, and what quantity comes out? - Which part of the charge is paid only once?

Solution

1. The input $3$ means the book is three days late. 2. Calculate $L(3)=2\cdot3+1=7$. The total charge is $\$7$.

Answer

$L(3)=7$ means a book returned three days late has a charge of $\$7$.

Function notation and evaluation 7

Consider the functions $f(x) = 4x - 2$, $g(x) = \frac{8}{x}$, and $h(x) = \frac{x}{2}$. Determine whether each graph crosses the $y$-axis. For each graph that does, give the $y$-intercept. For each graph that does not, explain why.

Hints

- What is the $x$-coordinate of every point on the $y$-axis? - Substitute that input into each function. - Check whether any substitution creates division by zero.

Solution

1. Substitute $x = 0$ into $f$: $f(0) = 4(0) - 2 = -2$. Therefore, the graph of $f$ crosses the $y$-axis at $(0, -2)$. 2. Substituting $x = 0$ into $g$ would give $g(0) = \frac{8}{0}$, which is undefined. Therefore, $0$ is not in the domain of $g$, and its graph does not cross the $y$-axis. 3. Substitute $x = 0$ into $h$: $h(0) = \frac{0}{2} = 0$. Therefore, the graph of $h$ crosses the $y$-axis at $(0, 0)$.

Answer

$f$: The graph crosses the $y$-axis at $(0, -2)$. $g$: The graph does not cross the $y$-axis because $g(0)$ is undefined. $h$: The graph crosses the $y$-axis at $(0, 0)$.

Function notation and evaluation 8

The function is $T(x)=x^2-3x+1$. Find $T(-3)$, $T(0.5)$, and $T(2)$.

Hints

- Use parentheses when substituting a negative input. - Evaluate exponents before multiplication and addition. - Track the sign of each term.

Solution

1. $T(-3)=(-3)^2-3\cdot(-3)+1=9+9+1=19$. 2. $T(0.5)=(0.5)^2-3\cdot0.5+1=0.25-1.5+1=-0.25$. 3. $T(2)=2^2-3\cdot2+1=4-6+1=-1$.

Answer

$T(-3)=19$ $T(0.5)=-0.25$ $T(2)=-1$

Function notation and evaluation 9

The function $T(x)=x^2+5$ is given. a) Find $T(1)$, $T(2)$, $T(3)$, $T(4)$, and $T(5)$. b) Which of these inputs gives the greatest function value?

Hints

- Substitute each listed input for $x$ in the function rule. - Keep the input and its output paired as you work. - For part b), compare the outputs after you have evaluated all five inputs.

Solution

1. Substitute each input into $T(x)=x^2+5$: $T(1)=6$, $T(2)=9$, $T(3)=14$, $T(4)=21$, and $T(5)=30$. 2. Compare the five outputs. The greatest is $30$, which occurs at input $x=5$.

Answer

a) $T(1)=6$, $T(2)=9$, $T(3)=14$, $T(4)=21$, and $T(5)=30$ b) $x=5$

Function notation and evaluation 10

The quadratic function $f$ is defined by $f(x)=x^2-3.5$. Determine whether each point lies on, above, or below the graph of $f$: $P_1(2, 0.5)$, $P_2(-1.5, -1.25)$, $P_3(0.5, -3.75)$, and $P_4(-1, 1)$. Show calculations.

Hints

- Evaluate the function at each point's x-coordinate. - Compare the calculated function value with the point's y-coordinate. - What does it mean when the point's y-coordinate is greater than or less than the function value?

Solution

1. For $P_1$, $f(2)=2^2-3.5=0.5$. This equals the point's y-coordinate, so $P_1$ lies on the graph. 2. For $P_2$, $f(-1.5)=(-1.5)^2-3.5=2.25-3.5=-1.25$. This equals the point's y-coordinate, so $P_2$ lies on the graph. 3. For $P_3$, $f(0.5)=(0.5)^2-3.5=0.25-3.5=-3.25$. Since $-3.75<-3.25$, $P_3$ lies below the graph. 4. For $P_4$, $f(-1)=(-1)^2-3.5=1-3.5=-2.5$. Since $1>-2.5$, $P_4$ lies above the graph.

Answer

$P_1$: on the graph $P_2$: on the graph $P_3$: below the graph $P_4$: above the graph

Function notation and evaluation 11

Let $f(x)=\frac{1}{4}x^2$. a) Determine algebraically whether $P(-2, 1)$ lies on the graph. b) Find all x-values for which $f(x)=16$. c) Evaluate $f(-3)$.

Hints

- Substitute the point's x-coordinate into the function and compare the result with its y-coordinate. - For part b), first isolate the squared expression before considering which real inputs can produce the target output. - Function notation $f(-3)$ means to substitute $-3$ for $x$.

Solution

1. Evaluate the function at $x=-2$: $f(-2)=\frac{1}{4}(-2)^2=1$. Therefore, $P$ lies on the graph. 2. Solve $\frac{1}{4}x^2=16$. Multiplying by $4$ gives $x^2=64$, so $x=-8$ or $x=8$. 3. $f(-3)=\frac{1}{4}(-3)^2=\frac{9}{4}=2.25$.

Answer

a) Yes, because $f(-2)=1$. b) $x=-8$ and $x=8$ c) $2.25$

Function notation and evaluation 12

Let $f(x)=5x^4-3x^3+2x^2-x+4$ and $g(x)=8x^3+4x^2-2x+1$. a) Evaluate $f(-1)$. b) Evaluate $g\left(-\frac{1}{2}\right)$.

Hints

- Read the function name carefully before substituting the requested input. - Use parentheses around a negative input before evaluating powers. - Evaluate powers before combining the remaining terms.

Solution

1. $f(-1)=5\cdot(-1)^4-3\cdot(-1)^3+2\cdot(-1)^2-(-1)+4=15$. 2. $g\left(-\frac{1}{2}\right)=8\cdot\left(-\frac{1}{2}\right)^3+4\cdot\left(-\frac{1}{2}\right)^2-2\cdot\left(-\frac{1}{2}\right)+1=2$.

Answer

a) $15$ b) $2$

Function notation and evaluation 13

Let $f(x)=4x^2-3x+12$, $g(a)=2a^3+5a-8$, and $h(y)=(y+4)(2y-1)$. a) Evaluate $f(5)$. b) Evaluate $g(3)$. c) Evaluate $h(6)$.

Hints

- Match each requested value with the correct function rule. - Substitute the input before evaluating powers and products. - In part c), evaluate each factor before multiplying.

Solution

1. $f(5)=4\cdot5^2-3\cdot5+12=97$. 2. $g(3)=2\cdot3^3+5\cdot3-8=61$. 3. $h(6)=(6+4)(2\cdot6-1)=110$.

Answer

a) $97$ b) $61$ c) $110$

Function notation and evaluation 14

The function is $T(n)=6n^3+4n+7$. Evaluate it for each input. a) $n=10$ b) $n=2$ c) $n=0.5$

Hints

- Cube the input before multiplying by $6$. - The exponent applies only to $n$. - Calculate carefully with the decimal input.

Solution

1. $T(10)=6\cdot10^3+4\cdot10+7=6000+40+7=6047$. 2. $T(2)=6\cdot2^3+4\cdot2+7=48+8+7=63$. 3. $T(0.5)=6\cdot0.5^3+4\cdot0.5+7=0.75+2+7=9.75$.

Answer

a) $6047$ b) $63$ c) $9.75$

Function notation and evaluation 15

Let $f(x)=x^3-4x$. Determine by substitution which of these points lie on the graph of $f$: $A(3, 15)$, $B(-2, 0)$, $C(1, 3)$, and $D(-1, 3)$.

Hints

- A point $(x, y)$ lies on a function's graph when substituting its x-coordinate gives its y-coordinate. - Substitute each x-coordinate into $f(x)$. - Pay close attention to parentheses and signs when evaluating powers of negative numbers.

Solution

1. For $A$, $f(3)=3^3-4\cdot 3=27-12=15$, which matches the y-coordinate. Therefore, $A$ lies on the graph. 2. For $B$, $f(-2)=(-2)^3-4\cdot(-2)=-8+8=0$, which matches the y-coordinate. Therefore, $B$ lies on the graph. 3. For $C$, $f(1)=1^3-4\cdot 1=-3$, which does not equal $3$. Therefore, $C$ does not lie on the graph. 4. For $D$, $f(-1)=(-1)^3-4\cdot(-1)=-1+4=3$, which matches the y-coordinate. Therefore, $D$ lies on the graph.

Answer

Points $A$, $B$, and $D$ lie on the graph. Point $C$ does not.

Function notation and evaluation 16

The following points lie on the graph of $f(x)=x^5$. Find each missing coordinate. a) $A(3,y_A)$ b) $B(x_B,-32)$ c) $C(-0.1,y_C)$ d) $D(x_D,0)$ e) $E(0.5,y_E)$

Hints

- A point on the graph must satisfy $y=x^5$. - When the output is known and the input is missing, solve the fifth-power equation for the input. - Track the sign of a negative number raised to an odd power. - You may rewrite decimals as fractions if that makes a power easier to evaluate.

Solution

1. For $A$, $y_A=3^5=243$, so $A(3,243)$. 2. For $B$, solve $x_B^5=-32$. Thus, $x_B=\sqrt[5]{-32}=-2$, so $B(-2,-32)$. 3. For $C$, $y_C=(-0.1)^5=-0.00001$, so $C(-0.1,-0.00001)$. 4. For $D$, $x_D^5=0$, so $x_D=0$ and $D(0,0)$. 5. For $E$, $y_E=(0.5)^5=0.03125$, so $E(0.5,0.03125)$.

Answer

a) $A(3,243)$ b) $B(-2,-32)$ c) $C(-0.1,-0.00001)$ d) $D(0,0)$ e) $E(0.5,0.03125)$

Function notation and evaluation 17

Let $f(x)=1.2746x+3.918$. A student rounds the coefficients before calculating and uses $g(x)=1.3x+3.9$ instead. a) Find $f(85)$ and $g(85)$. b) How much does the early rounding change the output?

Hints

- Use the same input in both rules. - Keep the given decimal places when evaluating the first rule. - Compare the two outputs after calculating each one.

Solution

1. With the original rule, $f(85)=1.2746\cdot85+3.918=112.259$. 2. With the rounded rule, $g(85)=1.3\cdot85+3.9=114.4$. 3. The rounded rule gives $114.4-112.259=2.141$ more. Rounding the coefficients before substituting changes the result.

Answer

a) $f(85)=112.259$; $g(85)=114.4$. b) The rounded rule gives an output $2.141$ higher.

Function notation and evaluation 18

The graph shows a function $f$. a) Find $f(-2)$. b) Find $f(4)$. c) Write the ordered pair on the graph that represents the value in part b.

Hints

- Start at the requested x-value and move vertically until you reach the graph. - The y-coordinate of that point is the function value. - Keep the same input when you write the corresponding ordered pair.

Solution

1. At $x=-2$, the graph has y-coordinate $5$, so $f(-2)=5$. 2. At $x=4$, the graph has y-coordinate $-1$, so $f(4)=-1$. 3. Therefore, the corresponding point is $(4,-1)$.

Answer

a) $5$ b) $-1$ c) $(4,-1)$

Function notation and evaluation 19

A parking garage models the charge for $h$ hours by $C(h)=4+2.5h$, where $0\le h\le8$. The output $C(h)$ is measured in dollars. a) Explain what the statement $C(3)=11.50$ means in this context. b) Find $C(6)$ and interpret the result. c) State the units of the input and the output of $C$.

Hints

- Read $C(h)$ as the cost associated with an input of $h$ hours. - For part b, substitute the requested number of hours for the input variable. - Use the context, not just the formula, to attach units to each quantity.

Solution

1. The statement $C(3)=11.50$ means that parking for $3$ hours costs $\$11.50$. 2. Evaluate $C(6)=4+2.5(6)=19$, so parking for $6$ hours costs $\$19$. 3. The input is time in hours, and the output is cost in dollars.

Answer

a) Parking for $3$ hours costs $\$11.50$. b) $C(6)=19$, so parking for $6$ hours costs $\$19$. c) Input: hours; output: dollars

Function notation and evaluation 20

For $f(x)=2x-1$, a student says, “$f(2)+f(3)=f(5)$ because $2+3=5$.” Check the claim and explain the mistake.

Hints

- Find the two outputs separately before adding them. - Now use the sum as one input. - What part of the rule is counted twice in the student's calculation?

Solution

1. Evaluate separately: $f(2)=3$ and $f(3)=5$, so $f(2)+f(3)=8$. 2. Evaluate the sum as one input: $f(5)=2\cdot5-1=9$. 3. The outputs do not add in the same way as the inputs, because the function includes a fixed subtraction of $1$.

Answer

The claim is false: $f(2)+f(3)=8$, but $f(5)=9$. Adding inputs does not always add their outputs.

Function notation and evaluation 21

The function $T(x)=1.5x^2-4.5$ is given. a) Find $T(-2)$, $T(-1.5)$, $T(-0.5)$, $T(0.5)$, $T(1.5)$, and $T(2)$. b) Identify the pairs of opposite inputs that have equal outputs. Explain why this happens from the function rule.

Hints

- Evaluate the function at each listed input before comparing outputs. - Compare inputs that have the same absolute value. - What happens to the sign when a number is squared?

Solution

1. Evaluate the function: $T(-2)=1.5$, $T(-1.5)=-1.125$, $T(-0.5)=-4.125$, $T(0.5)=-4.125$, $T(1.5)=-1.125$, and $T(2)=1.5$. 2. The equal-output pairs are $-2$ and $2$, $-1.5$ and $1.5$, and $-0.5$ and $0.5$. 3. Opposite inputs have the same square, so $T(-x)=1.5(-x)^2-4.5=1.5x^2-4.5=T(x)$.

Answer

a) $T(-2)=1.5$, $T(-1.5)=-1.125$, $T(-0.5)=-4.125$, $T(0.5)=-4.125$, $T(1.5)=-1.125$, and $T(2)=1.5$ b) The pairs are $(-2,2)$, $(-1.5,1.5)$, and $(-0.5,0.5)$. Opposite inputs give equal outputs because $(-x)^2=x^2$.

Function notation and evaluation 22

Evaluate each function for the given inputs. a) $T(y)=\frac{y^2+1}{y+2}$ for $y=4$ and $y=\frac{1}{2}$ b) $S(n)=n^3-2n^2+5$ for $n=3$ and $n=10$

Hints

- For a rational expression, evaluate the numerator and denominator separately. - Apply exponents before multiplication and subtraction. - Simplify each final fraction.

Solution

1. $T(4)=\frac{4^2+1}{4+2}=\frac{17}{6}=2\frac{5}{6}$. 2. $T\left(\frac{1}{2}\right)=\frac{\left(\frac{1}{2}\right)^2+1}{\frac{1}{2}+2}=\frac{\frac{5}{4}}{\frac{5}{2}}=\frac{1}{2}$. 3. $S(3)=3^3-2\cdot3^2+5=27-18+5=14$. 4. $S(10)=10^3-2\cdot10^2+5=1000-200+5=805$.

Answer

a) $T(4)=2\frac{5}{6}$; $T\left(\frac{1}{2}\right)=\frac{1}{2}$ b) $S(3)=14$; $S(10)=805$

Function notation and evaluation 23

The function is $f(x)=\frac{x^2+2x}{x+1.5}$. Complete the table, including the intermediate rows for $x^2$ and $2x$. Round each value of $f(x)$ to the nearest hundredth. <table><tbody><tr><td>$x$</td><td>$0.2$</td><td>$0.4$</td><td>$0.6$</td><td>$0.8$</td><td>$1.0$</td><td>$1.2$</td></tr><tr><td>$x^2$</td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td>$2x$</td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td>$f(x)$</td><td></td><td></td><td></td><td></td><td></td><td></td></tr></tbody></table>

Hints

- Evaluate the pieces of the numerator for each input before forming the quotient. - Keep each intermediate value in the same column as its input. - Round only the final function value in each column.

Solution

1. Square each input to obtain $0.04$, $0.16$, $0.36$, $0.64$, $1.00$, and $1.44$. 2. Double each input to obtain $0.4$, $0.8$, $1.2$, $1.6$, $2.0$, and $2.4$. 3. Substitute each input into $f(x)=\frac{x^2+2x}{x+1.5}$ and round only the final quotient: $0.26$, $0.51$, $0.74$, $0.97$, $1.20$, and $1.42$.

Answer

Function notation and evaluation 24

Let $f(x)=4x-3$. Determine whether each equation is an identity for all real values of its variable. Justify each answer by simplifying both sides. a) $f(2a)=2f(a)$ b) $f(x+1)=f(x)+4$ c) $f(x^2)=(f(x))^2$

Hints

- Replace the input variable with the entire expression inside the function parentheses. - Use parentheses when multiplying or squaring a complete function expression. - An identity must hold for every allowed value, not just one value.

Solution

1. $f(2a)=8a-3$, while $2f(a)=2(4a-3)=8a-6$. The expressions are not identical, so a is false. 2. $f(x+1)=4(x+1)-3=4x+1$, and $f(x)+4=(4x-3)+4=4x+1$. The expressions are identical, so b is true. 3. $f(x^2)=4x^2-3$, while $(f(x))^2=(4x-3)^2=16x^2-24x+9$. These expressions are not identical, so c is false. They are equal only when $x=1$.

Answer

a) False. $f(2a)=8a-3$, while $2f(a)=8a-6$, so the two sides are not identical. b) True. $f(x+1)=4x+1$ and $f(x)+4=4x+1$. c) False. $f(x^2)=4x^2-3$, while $(f(x))^2=(4x-3)^2$, so the expressions are not identical.

Function notation and evaluation 25

Let $f(x)=x^4$ and $g(x)=x^6$. a) Evaluate both functions at $x=0.5$ and $x=2$. b) Find $f(-3)$ and $g(-1.5)$. c) Evaluate $g(\sqrt{2})$. d) Simplify $f(2a)$ and $g(a^2)$.

Hints

- Even powers make negative inputs produce nonnegative outputs. - Use the power of a power property. - Relate $(\sqrt{2})^2$ to $2$. - When substituting $2a$, raise the entire product to the fourth power.

Solution

1. $f(0.5)=(0.5)^4=0.0625$ and $g(0.5)=(0.5)^6=0.015625$. 2. $f(2)=2^4=16$ and $g(2)=2^6=64$. 3. $f(-3)=(-3)^4=81$ and $g(-1.5)=(-1.5)^6=11.390625$. 4. $g(\sqrt{2})=(\sqrt{2})^6=((\sqrt{2})^2)^3=2^3=8$. 5. $f(2a)=(2a)^4=16a^4$, and $g(a^2)=(a^2)^6=a^{12}$.

Answer

a) $f(0.5)=0.0625$, $g(0.5)=0.015625$, $f(2)=16$, and $g(2)=64$ b) $f(-3)=81$ and $g(-1.5)=11.390625$ c) $g(\sqrt{2})=8$ d) $f(2a)=16a^4$ and $g(a^2)=a^{12}$

Function notation and evaluation 26

Let $f(x)=x^5$. a) Evaluate $f(x)$ for $x\in\{-2,-1,0,1,2\}$. b) Find $f(0.2)$ and $f(-0.4)$. c) For which real inputs is $f(x)$ positive? Explain. d) Simplify $f(-a)$ and $f(2a)$.

Hints

- An odd power keeps the sign of a nonzero input. - Look at the sign of the result, not just its size. - When the input is a product, raise the whole product to the fifth power.

Solution

1. $f(-2)=-32$, $f(-1)=-1$, $f(0)=0$, $f(1)=1$, and $f(2)=32$. 2. $f(0.2)=(0.2)^5=0.00032$, and $f(-0.4)=(-0.4)^5=-0.01024$. 3. A positive number to the fifth power is positive; zero stays zero and a negative number gives a negative fifth power. Thus $f(x)>0$ exactly when $x>0$. 4. $f(-a)=(-a)^5=-a^5$ and $f(2a)=(2a)^5=32a^5$.

Answer

a) $f(-2)=-32$, $f(-1)=-1$, $f(0)=0$, $f(1)=1$, $f(2)=32$. b) $f(0.2)=0.00032$, $f(-0.4)=-0.01024$. c) $x>0$. d) $f(-a)=-a^5$, $f(2a)=32a^5$.

Function notation and evaluation 27

The graph of $f(x)=ax^4-2x^2$ passes through $P(2, 4)$. a) Find the value of $a$. b) Determine whether $Q(-1, -1.25)$ also lies on the graph.

Hints

- Substitute the coordinates of $P$ into the equation to create an equation in $a$. - After finding $a$, evaluate the function at the x-coordinate of $Q$. - An even power of a negative number is positive.

Solution

1. Substitute $P(2, 4)$: $4=a\cdot 2^4-2\cdot 2^2$. 2. Simplify: $4=16a-8$, so $12=16a$ and $a=\frac{3}{4}=0.75$. 3. The function is $f(x)=0.75x^4-2x^2$. 4. Evaluate at $x=-1$: $f(-1)=0.75(-1)^4-2(-1)^2=0.75-2=-1.25$. 5. This equals the y-coordinate of $Q$, so $Q$ lies on the graph.

Answer

a) $a=0.75$ b) Yes. $Q(-1, -1.25)$ lies on the graph.

Function notation and evaluation 28

The points below lie on the graph of $g(x)=x^6$. Find each missing coordinate. Remember that solving for an input may give two values. a) $P_1(-2,y_1)$ b) $P_2(x_2,1)$ c) $P_3(0.5,y_3)$ d) $P_4(x_4,1{,}000{,}000)$ e) $P_5(x_5,0)$

Hints

- An even power gives the same output for opposite nonzero inputs. - When a sixth power equals a positive number, check whether both a positive and a negative input work. - Rewrite $1{,}000{,}000$ as a power of $10$. - Consider separately what happens when the output is zero.

Solution

1. For $P_1$, $y_1=(-2)^6=64$. 2. For $P_2$, solve $x_2^6=1$. The real solutions are $x_2=1$ and $x_2=-1$. 3. For $P_3$, $y_3=(0.5)^6=\frac{1}{64}=0.015625$. 4. For $P_4$, solve $x_4^6=1{,}000{,}000=10^6$. The real solutions are $x_4=10$ and $x_4=-10$. 5. For $P_5$, $x_5^6=0$, so $x_5=0$.

Answer

a) $P_1(-2,64)$ b) $P_2(1,1)$ or $P_2(-1,1)$ c) $P_3(0.5,0.015625)$ d) $P_4(10,1{,}000{,}000)$ or $P_4(-10,1{,}000{,}000)$ e) $P_5(0,0)$

Function notation and evaluation 29

The table defines a function $q$. <table><tr><td>$x$</td><td>$-4$</td><td>$-1$</td><td>$2$</td><td>$5$</td><td>$7$</td></tr><tr><td>$q(x)$</td><td>$6$</td><td>$1$</td><td>$6$</td><td>$-3$</td><td>$0$</td></tr></table> a) Find every input $a$ for which $q(a)=6$. b) Write the ordered pair represented by $q(5)=-3$. c) Explain why having two different inputs with output $6$ does not violate the definition of a function.

Hints

- Reverse function notation asks you to start from an output and locate the input or inputs that produce it. - Read across the output row first, then trace matching entries back to their inputs. - The definition of a function restricts the number of outputs for one input, not the number of inputs that may share one output.

Solution

1. In the output row, $6$ occurs under inputs $-4$ and $2$, so $q(-4)=6$ and $q(2)=6$. 2. The statement $q(5)=-3$ corresponds to the ordered pair $(5, -3)$. 3. A function requires each input to have exactly one output. Different inputs are allowed to share the same output, so the repeated output $6$ does not violate the function definition.

Answer

a) $a=-4$ and $a=2$ b) $(5, -3)$ c) A function may assign the same output to different inputs; each individual input still has exactly one output.

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Algebra I math worksheets

Function notation and evaluation

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