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Piecewise Functions: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The function machine

Let $f(x)=x+2$ for $x<0$ , and $f(x)=3-x$ for $x\ge0$ . Find $f(f(-1))$ .
1. Hint 1
  The first output becomes the input for the second application.
2. Hint 2
  Choose a piece separately for each input rather than keeping the first choice.
Answer

$2$ .

Full solution

The input $-1$ is negative, so
$f(-1)=-1+2$
giving $1$ .
That output is nonnegative, so the other piece applies next:

$f(1)=3-1$
Thus the requested value is $2$ .

Answer

$2$ .

Key idea

Each application of a piecewise function chooses its piece from its own current input.
Problem 1: Mark complete
Problem 2 The hidden threshold

A function $f$ returns $5$ when $x<t$ and $-1$ when $x\ge t$ , where $t$ is real. The records say $f(2)=5$ and $f(3)=-1$ . Find all possible $t$ .
1. Hint 1
  Each recorded output tells you which condition its input satisfies.
2. Hint 2
  The input $2$ must lie below the threshold, while $3$ must be at or above it.
Answer

$2<t\le3$ .

Full solution

The output $5$ at input $2$ requires $2<t$ .
The output $-1$ at input $3$ requires $3\ge t$ .

Together the conditions give
$2<t\le3$
Every threshold in this interval assigns both inputs their recorded outputs.

Answer

$2<t\le3$ .

Key idea

Recorded outputs of constant pieces can constrain an unknown switching threshold.
Problem 2: Mark complete
Problem 3 Which piece owns the input

Let $f(x)=-x-1$ for $x\le-3$ , let $f(x)=x^2-2$ for $-3<x\le1$ , and let $f(x)=2x+3$ for $x>1$ . Find $f(-5)$ , $f(-3)$ , $f(0)$ , $f(1)$ , and $f(4)$ .
1. Hint 1
  Settle which condition an input satisfies before substituting it into any formula.
2. Hint 2
  At each boundary, exactly one condition carries an equality sign, and that piece owns the input.
Answer

$f(-5)=4$ , $f(-3)=2$ , $f(0)=-2$ , $f(1)=-1$ , $f(4)=11$ .

Full solution

Both $-5$ and $-3$ satisfy $x\le-3$ , so the first piece $-x-1$ applies to each of them.
Negating the input turns $-5$ into $5$ and $-3$ into $3$ , so the outputs are $5-1$ and $3-1$ :
$\begin{aligned}f(-5)&=4\\f(-3)&=2\end{aligned}$

Both $0$ and $1$ satisfy $-3<x\le1$ , so the middle piece $x^2-2$ applies.
Their squares are $0$ and $1$ , so the outputs are $0-2$ and $1-2$ :
$\begin{aligned}f(0)&=-2\\f(1)&=-1\end{aligned}$
The boundary $1$ lands here because this condition includes it, while the boundary $-3$ went to the first piece for the same reason.

The input $4$ satisfies $x>1$ , so the last piece gives $2(4)+3$ , that is
$f(4)=11$

Answer

$f(-5)=4$ , $f(-3)=2$ , $f(0)=-2$ , $f(1)=-1$ , $f(4)=11$ .

Key idea

Reading the condition first, and noticing which side of a boundary carries the equality sign, decides every value of a piecewise function.
Problem 3: Mark complete
Problem 4 The cropped graph

Let $f(x)=2-x$ for $x\le3$ , and $f(x)=(x-3)^2-1$ for $x>3$ . On the axes in the figure, sketch only the part with $1\le x\le5$ . State the range of the part you draw and whether the two pieces connect at $x=3$ .

A blank grid for $0 \le x \le 6$ and $-2 \le y \le 4$ .

Text description of this figure

A blank coordinate grid with equal unit lengths on both axes. The horizontal x-axis runs from 0 to 6 and the vertical y-axis runs from negative two to four, with gridlines, tick marks and number labels at every whole number, the origin labeled 0, an arrowhead at the right end of the x-axis and an arrowhead at the top of the y-axis. Nothing is plotted on the grid: no points, segments, curves or equations.
1. Hint 1
  Restrict both formulas to the requested drawing interval before choosing endpoints or outputs.
2. Hint 2
  Compare the two formulas at the switching input, and find the lowest and highest drawn heights.
Answer

Range $-1\le y\le3$ ; the pieces connect at $(3,-1)$ .

Full solution

The line piece runs from $(1,1)$ to $(3,-1)$ , including both endpoints.
The quadratic piece runs from just beyond $(3,-1)$ to $(5,3)$ , including the outer endpoint but not its input $3$ .

Both formulas have boundary height $-1$ .
The first includes that point, so the pieces connect.
The line gives heights from $-1$ to $1$ , and the quadratic gives heights greater than $-1$ through $3$ .

The combined range is
$-1\le y\le3$
Draw no portion outside the requested input interval.

Answer

Range $-1\le y\le3$ ; the pieces connect at $(3,-1)$ .

Key idea

Restrict each piece to the part being drawn before reading off the heights it actually reaches.
Problem 4: Mark complete
Problem 5 The three segments

The figure shows all of a piecewise function. Write its rule with one formula for each drawn segment, and state its domain and range.

The complete graph of the function, drawn as three segments.

Text description of this figure

A square coordinate grid with equal unit lengths on both axes, gridlines, tick marks and number labels at every whole number, the horizontal x-axis running from negative five to five and the vertical y-axis from negative two to six, the origin labeled 0 and arrowheads on both ends of each axis. Three straight segments are drawn. The first falls from a filled dot at the point (negative 4, 2) down to a hollow dot at the point (negative 1, negative 1). The second is horizontal at height 3, from a filled dot at the point (negative 1, 3) across to a filled dot at the point (2, 3). The third rises from that same filled dot at the point (2, 3) up to a filled dot at the point (4, 5). Each of those five endpoints is labeled with its coordinates. Nothing else is drawn: no formulas, no domain or range labels and no graph outside these three segments.
1. Hint 1
  Each straight segment is fixed by its two labeled endpoints: get its slope from them, then its rule.
2. Hint 2
  Combine the input intervals for the domain and the attained heights for the range.
Answer

$f(x)=-x-2$ on $[-4,-1)$ ; $f(x)=3$ on $[-1,2]$ ; $f(x)=x+1$ on $(2,4]$ , or equivalently $f(x)=3$ on $[-1,2)$ and $f(x)=x+1$ on $[2,4]$ . Domain $[-4,4]$ ; range $(-1,2]\cup[3,5]$ .

Full solution

The sloping left segment passes through $(-4,2)$ and approaches $(-1,-1)$ , so its slope is $-1$ and its rule is $-x-2$ on
$-4\le x<-1.$
Its heights satisfy $-1<y\le2$ .

The middle segment is $y=3$ on
$-1\le x\le2.$
The right segment has slope $1$ and rule $x+1$ on $2<x\le4$ , reaching heights $3<y\le5$ .

The conditions cover
$-4\le x\le4$
Combining the attained heights gives
$(-1,2]\cup[3,5]$
The height $3$ is included by the middle piece.

The point $(2,3)$ may instead be assigned to the right piece and excluded from the middle piece; that choice gives the same function.

Answer

$f(x)=-x-2$ on $[-4,-1)$ ; $f(x)=3$ on $[-1,2]$ ; $f(x)=x+1$ on $(2,4]$ , or equivalently $f(x)=3$ on $[-1,2)$ and $f(x)=x+1$ on $[2,4]$ . Domain $[-4,4]$ ; range $(-1,2]\cup[3,5]$ .

Key idea

A piecewise graph supplies separate equations and endpoint conditions whose output ranges must be combined.
Problem 5: Mark complete
Problem 6 The limited display

A display takes any real reading $x$ . It reports $-2$ if $x<-2$ , reports $3$ if $x>3$ , and otherwise reports the reading unchanged. Write a piecewise rule, find the displayed values for inputs $-5$ , $1$ , and $4$ , and state the range.
1. Hint 1
  The unchanged region includes the two boundary readings.
2. Hint 2
  Use a separate condition for each constant output and for the middle identity rule.
Answer

$f(x)=-2$ if $x<-2$ , $x$ if $-2\le x\le3$ , and $3$ if $x>3$ ; outputs $-2,1,3$ ; range $[-2,3]$ .

Full solution

The instructions translate directly into three mutually exclusive conditions: $x<-2$ gives $-2$ , $-2\le x\le3$ gives $x$ , and $x>3$ gives $3$ .

The three requested inputs belong to the left, middle, and right regions respectively, so
$\begin{aligned}f(-5)&=-2\\f(1)&=1\\f(4)&=3\end{aligned}$

The middle piece reaches every value from $-2$ to $3$ , including both ends.
The outer pieces add no further heights, so the range is $[-2,3]$ .

Answer

$f(x)=-2$ if $x<-2$ , $x$ if $-2\le x\le3$ , and $3$ if $x>3$ ; outputs $-2,1,3$ ; range $[-2,3]$ .

Key idea

A display that caps extreme readings is described by constant outer pieces and an unchanged middle piece.
Problem 6: Mark complete
Problem 7 The step labels

A function equals $-2$ on $-2<x\le-1$ , $-1$ on $-1<x\le0$ , $0$ on $0<x\le1$ , and $1$ on $1<x\le2$ . Write one formula using a ceiling and state its domain. Explain how its endpoint ownership agrees with that formula.
1. Hint 1
  Each interval includes its right integer endpoint and excludes its left one.
2. Hint 2
  Compare each output with the ceiling of an input in its interval.
Answer

$f(x)=\lceil x\rceil-1$ ; domain $-2<x\le2$ .

Full solution

On the four intervals, the ceiling values are $-1$ , $0$ , $1$ , and $2$ .
Subtracting $1$ gives exactly the four specified outputs.

Thus
$f(x)=\lceil x\rceil-1$
on $-2<x\le2$ .
Ceiling pieces own their right endpoints, matching every stated interval, and the domain must remain restricted as given.

Answer

$f(x)=\lceil x\rceil-1$ ; domain $-2<x\le2$ .

Key idea

A finite piecewise staircase can match a shifted ceiling on its stated domain.
Problem 7: Mark complete
Problem 8 The alternative rule

Let $g(x)=x-1$ for $x\ge1$ , and $g(x)=1-x$ for $x<1$ . Tariq claims that $g(x)=|x|-1$ for every real input. Is he correct? Explain.
1. Hint 1
  A rule claimed for every input is refuted as soon as one input disagrees.
2. Hint 2
  Each piece of $g$ reports how far the input sits from $1$ , while the claimed rule starts from the distance from $0$ .
Answer

No; at $x=0$ the pieces give $1$ , while $|0|-1=-1$ .

Full solution

The input $0$ satisfies $x<1$ , so the second piece applies and
$g(0)=1$

The claimed rule gives $|0|-1=-1$ at that same input.
One disagreeing input settles the question, so the claim is false.

The pieces keep $x-1$ when it is nonnegative and negate it when it is negative, so $g(x)=|x-1|$ , the distance from $x$ to $1$ .
The claimed $|x|-1$ instead measures the distance from $x$ to $0$ and then lowers every height by one.

Answer

No; at $x=0$ the pieces give $1$ , while $|0|-1=-1$ .

Key idea

The number inside absolute value bars marks where the two sign pieces switch, and is not a change made to the outputs.
Problem 8: Mark complete
Problem 9 Ben's adjustment

A function agrees with $\lfloor x\rfloor$ at every real input except $0$ , where its value is changed to $-1/2$ . Ben claims this change joins the two floor steps on either side of $0$ . Is he correct? Explain using the nearby pieces.
1. Hint 1
  Ask what heights the graph actually takes just to the left and just to the right of zero.
2. Hint 2
  A step is a whole horizontal piece rather than a single point, so ask what one point at $x=0$ can and cannot fill.
Answer

No; the neighboring step heights remain $-1$ and $0$ .

Full solution

On $-1\le x<0$ , the unchanged rule gives $-1$ .
On $0<x<1$ , it gives $0$ .
These different heights remain separated, regardless of the assigned value at the single input $0$ .

The new point is
$(0,-1/2)$
It lies on neither neighboring horizontal step, so the graph still does not connect at zero.

Answer

No; the neighboring step heights remain $-1$ and $0$ .

Key idea

Changing one boundary output does not join neighboring constant pieces that approach different heights.
Problem 9: Mark complete
Problem 10 The studio rental fee

A studio charges a flat $b$ dollars for any rental of up to and including $3$ hours, and $4h-3$ dollars for a rental of $h$ hours when $h>3$ , where $b$ is a positive constant and $h>0$ . Find the value of $b$ that makes the cost graph connect at $h=3$ , and describe what happens to the cost at $3$ hours if the studio sets $b=10$ instead.
1. Hint 1
  Connecting at a boundary means the two pieces settle on the same height there.
2. Hint 2
  Work out the height the second formula approaches at $3$ hours, then compare it with the flat fee.
Answer

$b=9$ . With $b=10$ the cost is $10$ dollars at exactly $3$ hours but just above $9$ dollars for slightly longer rentals, a drop of about a dollar.

Full solution

The flat piece owns every rental time up to and including $3$ hours, so its height at the boundary is $b$ dollars.
The second formula approaches
$4(3)-3=9$
as the rental time comes down toward $3$ hours.

The two heights agree exactly when
$b=9$
so that fee joins the pieces and the cost graph connects at $3$ hours.

With $b=10$ there is a closed dot at $(3,10)$ and an open dot at $(3,9)$ , so the cost drops by about a dollar just past $3$ hours.
A customer who books a few minutes over $3$ hours then pays less than one who books exactly $3$ hours.

Answer

$b=9$ . With $b=10$ the cost is $10$ dollars at exactly $3$ hours but just above $9$ dollars for slightly longer rentals, a drop of about a dollar.

Key idea

Matching the two pieces at the threshold fixes the constant in a tiered price, and any other value leaves a step at the boundary.
Problem 10: Mark complete

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