Module 02: Functions, inputs, domains, and graphs

8 exercises, each with a complete solution. 1 figure.

On this page

A. Prerequisites and observable goals
B. The central idea: distinguish the rule from its output
C. Definitions and graph language
D. Why the input shift moves the graph
E. Recognize what the problem requests
F. Worked examples and a fading step
Computing bridge: Function notation and program inputs
G. Independent practice
H. Complete solutions: Module 02
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Module 01, especially parentheses and restrictions.

After

Read a function as an input-output rule, evaluate shifted inputs, identify domains and ranges, and translate graph transformations into coordinates.

B. The central idea: distinguish the rule from its output#

A real-valued function assigns exactly one real output to each input in its domain. The symbol $f$ names the rule; $f(x)$ names its output at input $x$ . The parentheses in $f(3)$ are function notation, not multiplication by a number $f$ . You can rename the input placeholder without changing the rule:

f(x)=2x^2-3x+4\quad\text{and}\quad f(t)=2t^2-3t+4.

In both cases, the instruction is “double the square of the input, subtract three times the input, then add four.” Consequently,

f(\boxed{\phantom{x}})=2(\boxed{\phantom{x}})^2-3(\boxed{\phantom{x}})+4.

Substituting means filling every box with the same complete input. When the input is $x+h$ , the whole expression goes into every box. It is not an instruction to add $h$ after computing $f(x)$ .

The rule and one output. The curve is the whole rule

f(x)=2x^2-3x+4

. Evaluating at the input

2

means going up from

2

to the curve and across to the output axis:

f(2)=6

. The point

(2,f(2))

is one point of the graph;

f

is all of them.

C. Definitions and graph language#

The domain is the set of allowed inputs. The range is the set of outputs actually attained. A graph contains ordered pairs $(x,f(x))$ ; its horizontal coordinate is an input and its vertical coordinate an output. A vertical line cannot meet a function graph in two distinct points. A horizontal line may: different inputs can have the same output.

For a formula with no prescribed domain, use its natural real domain: exclude denominator zeros, require even-root inputs to be nonnegative, and later require logarithm inputs to be positive. A contextual domain may be smaller. A time variable for a launched object is often restricted to the physical flight interval even though its polynomial formula accepts every real number.

For a point $(u,f(u))$ , the transformed graph

y=A f(B(x-c))+d,\qquad B\ne0,

contains the point $(c+u/B,\,Af(u)+d)$ . This coordinate rule explains horizontal changes rather than relying on a misleading “opposite sign” slogan. Absolute values, piecewise formulas, and excluded endpoints require attention to their definitions. A piecewise function is one rule specified by cases; choose the case using the input before evaluating.

D. Why the input shift moves the graph#

Suppose $g(x)=f(x-3)$ . To obtain the old output $f(2)$ , the new input must satisfy $x-3=2$ , hence $x=5$ . Thus the old graph point at input $2$ moves right by $3$ . The graph is not moving left simply because the formula contains a minus sign. Input arithmetic is evaluated before the function acts; output arithmetic happens afterward.

For $g(x)=f(2x)$ , reaching old input $u$ requires $x=u/2$ . All horizontal coordinates are halved. For $g(x)=2f(x)$ , horizontal coordinates stay fixed and heights double. These are different transformations. Graphing software can illustrate them, but a window can hide asymptotes, holes, or distant behavior; use algebra to identify those features. Definitions checked against standard function treatment [O1F].

E. Recognize what the problem requests#

If it asks for $f(a)$ , substitute. If it asks for $f(x+h)$ , substitute the whole expression and simplify only afterward. If it asks to solve $f(x)=k$ , find inputs that give the specified output. If it asks for a domain, list restrictions and intersect them; do not try to find a derivative. If it asks for range, consider which outputs are possible, using a graph, algebra, or known bounds.

F. Worked examples and a fading step#

Example 1: an input versus an output change. Let $f(x)=x^2-4x$ . Then

\begin{align*} f(x+h)&=(x+h)^2-4(x+h)\\ &=x^2+2xh+h^2-4x-4h,\\ f(x)+h&=x^2-4x+h. \end{align*}

Their difference is $2xh+h^2-5h$ , generally nonzero. At $x=2,h=1$ , the first is $f(3)=-3$ and the second is $f(2)+1=-3$ ; this accidental agreement at one input is not an identity. At $x=0,h=1$ , they are $-3$ and $1$ . A single counterexample disproves an all-input claim.

Example 2: a domain. For $g(x)=\sqrt{5-x}/(x+1)$ , require $x\le5$ and $x\ne-1$ . Therefore the domain is $(-\infty,-1)\cup(-1,5]$ . The square-root endpoint $5$ is included because the denominator is then $6$ , not zero.

Pause and finish. If $q(x)=|x|$ , where does $(2,2)$ move under $y=-3q(x+4)+1$ ? Solve $x+4=2$ , then transform the height. Completion: $x=-2$ , $y=-3(2)+1=-5$ , giving $(-2,-5)$ .

Computing bridge: Function notation and program inputs#

In a pure numerical function, replacing the input by $x+h$ is analogous to calling the same function with a new argument, not adding $h$ to its returned value. For $f(x)=x^2$ , $f(x+h)=(x+h)^2$ whereas $f(x)+h=x^2+h$ . They are different programs as well as different formulas. C01 makes this distinction with input values, returned values, and explicit state. A function with side effects requires additional reasoning: evaluating a call twice can change state, unlike substituting twice into a fixed mathematical function.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M02.01FSolution

For $f(x)=3x^2-2x+5$ , find $f(-2)$ and $f(x+h)$ without confusing the input with the output.

M02.02FSolution

For $g(x)=x^2+1$ , compare $g(2+3)$ with $g(2)+3$ . What does the comparison establish?

M02.03CSolution

Find the real domain of $\sqrt{x+2}/(x-4)$ .

M02.04CSolution

Let $p(x)=x+1$ for $x<0$ and $p(x)=x^2$ for $x\ge0$ . Find $p(-2),p(0),p(2)$ and all solutions of $p(x)=1$ .

M02.05CSolution

Give the domain and range of $y=2-\sqrt{x-1}$ . Explain both endpoints.

M02.06TSolution

A graph of $f$ contains $(-3,4)$ . Find the corresponding point on $g(x)=-2f(3(x-1))+5$ .

M02.07TSolution

Does the relation $x^2+y^2=9$ define all of $y$ as one function of $x$ ? Give two function branches and their domains.

M02.08SSolution

A data table contains $f(1)=4$ , $f(2)=7$ , $f(3)=12$ . Does it prove $f(x)=x^2+3$ for all real $x$ ? Construct a different polynomial agreeing with all three entries.

Optional small hints

For M02.06, solve $3(x-1)=-3$ . For M02.08, add a nonzero polynomial that vanishes at the three listed inputs.

H. Complete solutions: Module 02#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M02.01Show solution

Substitute into every input position

For the number $-2$ , write $f(-2)=3(-2)^2-2(-2)+5=12+4+5=\boxed{21}$ . Parentheses make the square apply to the entire negative input. For $x+h$ ,

f(x+h)=3(x+h)^2-2(x+h)+5 =\boxed{3x^2+6xh+3h^2-2x-2h+5}.

The last constant does not change because it was not an input slot.

Back to M02.01

M02.02Show solution

Two different operations

The first expression is $g(5)=25+1=26$ . The second is $(2^2+1)+3=8$ . Since $26\ne8$ , input addition cannot generally be moved outside a function. This is a counterexample to a proposed identity, not a statement that the two expressions can never agree for any function or input.

Back to M02.02

M02.03Show solution

Intersect restrictions

The radical requires $x+2\ge0$ , so $x\ge-2$ . The denominator excludes $4$ . Intersecting gives $\boxed{[-2,4)\cup(4,\infty)}$ . At $-2$ the numerator is zero and denominator is $-6$ , so the endpoint is allowed.

Back to M02.03

M02.04Show solution

Choose the correct branch first

Negative input $-2$ uses $x+1$ , giving $-1$ . The inputs $0,2$ use $x^2$ , giving $0,4$ . To solve $p(x)=1$ , the negative branch gives $x+1=1$ , or $x=0$ , but $0$ is not in that branch’s domain. The nonnegative branch gives $x^2=1$ , with candidates $\pm1$ , of which only $1$ belongs. Thus the solution is $\boxed{x=1}$ .

Back to M02.04

M02.05Show solution

Possible inputs and possible heights

The input must satisfy $x-1\ge0$ , so the domain is $[1,\infty)$ . The square root takes every value in $[0,\infty)$ . Subtracting those values from $2$ produces $(-\infty,2]$ . The maximum height $2$ occurs at $x=1$ ; there is no lowest height because the square root grows without bound.

Back to M02.05

M02.06Show solution

Transform coordinates, not just a formula

The old input is $-3$ . Set $3(x-1)=-3$ to get $x=0$ . The new height is $-2(4)+5=-3$ . Therefore the corresponding point is $\boxed{(0,-3)}$ . Direct verification: $g(0)=-2f(-3)+5=-3$ .

Back to M02.06

M02.07Show solution

A relation with two outputs

At $x=0$ , the relation allows $y=3$ and $y=-3$ , so the full circle fails the function requirement. Solving for $y$ gives branches $\boxed{y=\sqrt{9-x^2}}$ and $\boxed{y=-\sqrt{9-x^2}}$ . Both have domain $[-3,3]$ . Together they describe the circle; separately each assigns one output to every allowed input.

Back to M02.07

M02.08Show solution

Finite data do not determine a unique global rule

The polynomial $x^2+3$ matches the table, but so does

\boxed{q(x)=x^2+3+(x-1)(x-2)(x-3)}.

At each listed input, one of the three added factors is zero. At $x=0$ , however, $q(0)=3-6=-3$ , while $x^2+3=3$ . Thus the rules differ. Tables are evidence about listed inputs, not proofs of a global formula or a limit.

Back to M02.08

I. Retention and readiness#

Retrieve M01.02 and M01.06 alongside M02.01. Explain the difference among $f$ , $f(a)$ , and solving $f(x)=a$ . The correct explanation names a rule, an output, and an input-finding task, respectively. Advance when you can do M02.04 and a fresh domain problem without looking at solutions; otherwise repair input notation before using a difference quotient.