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Functions and Their Graphs: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

0 of 10 completed

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Problem 1 of 10

Problem 1 Changing the horizontal coordinate

Difficulty: 1 of 3 stars, Stretch

Start with the graph $y=|x|$ for all real $x$ . For a fixed real number $k$ , move every point $(x,y)$ to the point with coordinates
$u=x+ky,\qquad v=y.$
This rule changes the horizontal coordinate by an amount depending on the height.

Find every $k$ for which the resulting set is the graph of a function $v=F(u)$ . For every permitted $k$ , give the full domain and a formula for $F$ . Prove that every excluded value fails the vertical-line test.

Builds on Relations and Functions
1. Hint 1
  Follow the two rays $x\ge0$ and $x\le0$ separately.
2. Hint 2
  On those rays, respectively, $u=(1+k)x$ and $u=(1-k)x$ . Decide whether their images occupy opposite sides of the $u$ -axis or overlap.
Answer

Exactly $-1<k<1$ . The domain is $\mathbb R$ , and $F(u)=u/(1+k)$ for $u\ge0$ , while $F(u)=-u/(1-k)$ for $u\le0$ .

Full solution

On the right ray, $y=x$ and $u=(1+k)x$ with $x\ge0$ .
On the left ray, $y=-x$ and $u=(1-k)x$ with $x\le0$ .
If $-1<k<1$ , both coefficients are positive.
The right ray therefore gives exactly $u\ge0$ , and the left ray exactly $u\le0$ .
Each produces one height at each of its inputs, and the two heights agree at $u=0$ .
Solving for $x$ gives the stated formulas and domain.

If $k=1$ , the entire left ray has $u=0$ and varying heights.
If $k=-1$ , the right ray does.
Both fail the vertical-line test.

If $k>1$ , both rays produce every positive $u$ .
At such an input their heights are $u/(k+1)$ and $u/(k-1)$ , which differ.
If $k<-1$ , both produce every negative $u$ , with those same two formulas for heights; again the denominators differ and the heights are distinct.
Thus every remaining setting also fails.
An invertible point transformation of the plane need not preserve the property of being a function of its horizontal coordinate.

Answer

Exactly $-1<k<1$ . The domain is $\mathbb R$ , and $F(u)=u/(1+k)$ for $u\ge0$ , while $F(u)=-u/(1-k)$ for $u\le0$ .

Key idea

To test a transformed relation, track which horizontal inputs each branch covers, not just the movement of individual points.
Problem 1: Mark complete
Problem 2 A translation chosen by the input

Difficulty: 1 of 3 stars, Stretch

For a real number $a$ , define $f_a:\mathbb R\to\mathbb R$ by
$f_a(x)=\begin{cases}x+a&\text{if }x\text{ is rational},\\x-a&\text{if }x\text{ is irrational}.\end{cases}$
A rational number is a ratio of integers with nonzero denominator.

Find all $a$ for which $f_a$ is a bijection. For each permitted $a$ , give the inverse formula. For each excluded $a$ , exhibit both two distinct inputs with the same output and an output that is never attained.
1. Hint 1
  If $a$ is rational, adding or subtracting it preserves whether a number is rational.
2. Hint 2
  If $a$ is irrational, compare the inputs $0$ and $2a$ , and investigate whether $-a$ can be an output.
Answer

Exactly rational $a$ . Then $f_a^{-1}(y)=y-a$ for rational $y$ , and $f_a^{-1}(y)=y+a$ for irrational $y$ . For irrational $a$ , $0$ and $2a$ both map to $a$ , and $-a$ is missing.

Full solution

Suppose first that $a$ is rational.
Rational inputs remain rational after adding $a$ , and irrational inputs remain irrational after subtracting $a$ .
Thus outputs from the two branches cannot collide.
Each branch is a translation and hence injective within its input class.
Every rational $y$ is obtained from the rational input $y-a$ , and every irrational $y$ from the irrational input $y+a$ .
This proves bijectivity and gives the inverse.

Now suppose $a$ is irrational.
Then $2a$ is irrational and nonzero, so
$f_a(0)=a=f_a(2a).$
These are distinct inputs, disproving injectivity.

To obtain output $-a$ from the rational branch would require $x+a=-a$ , or $x=-2a$ , which is irrational and therefore not allowed on that branch.
To obtain it from the irrational branch would require $x-a=-a$ , or $x=0$ , which is rational and not allowed there.
Hence $-a$ is also absent from the range.
The same parameter change breaks both injectivity and surjectivity, but each failure needs its own verification.

Answer

Exactly rational $a$ . Then $f_a^{-1}(y)=y-a$ for rational $y$ , and $f_a^{-1}(y)=y+a$ for irrational $y$ . For irrational $a$ , $0$ and $2a$ both map to $a$ , and $-a$ is missing.

Key idea

A piecewise rule can be inverted branch by branch only after you know how the branches’ output sets interact.
Problem 2: Mark complete
Problem 3 Error after repeated application

Difficulty: 1 of 3 stars, Stretch

A function $f:[0,1]\to[0,1]$ satisfies $|f(x)-x|\le1/5$ for every $x\in[0,1]$ . Applying $f$ repeatedly $n$ times is denoted by $f^{\circ n}$ ; for example, $f^{\circ3}(x)=f(f(f(x)))$ .

For each positive integer $n$ , find the smallest number $B_n$ guaranteed to satisfy
$|f^{\circ n}(x)-x|\le B_n$
for every such function and every $x\in[0,1]$ . Prove the bound and give one explicit function that shows all your bounds are sharp. No continuity or monotonicity is assumed.
1. Hint 1
  Write the change after $n$ applications as a sum of $n$ consecutive changes.
2. Hint 2
  The output interval provides a second bound independent of $n$ . For sharpness, consider a function that moves right by $1/5$ until it reaches the endpoint.
Answer

$B_n=\min(n/5,1)$ . One function attaining every bound at $x=0$ is $f(x)=\min(x+1/5,1)$ .

Full solution

Let $x_0=x$ and $x_j=f(x_{j-1})$ for
$j=1,\ldots,n.$
Every $x_j$ remains in $[0,1]$ because that is the codomain of $f$ .
Thus the given error condition can be used at every step, not just the first.

By adding the consecutive changes and using the triangle inequality,
$|x_n-x_0|\le|x_n-x_{n-1}|+\cdots+|x_1-x_0|\le n/5.$
Also $x_n,x_0\in[0,1]$ , so
$|x_n-x_0|\le1.$
This proves the combined bound $\min(n/5,1)$ .

For sharpness, define
$f(x)=\min(x+1/5,1).$
It maps the interval into itself, and its change at each input lies between $0$ and $1/5$ .
Starting at $0$ , successive outputs are $1/5,2/5,3/5,4/5,1,1,\ldots$ .
Hence $f^{\circ n}(0)=\min(n/5,1)$ for every positive integer $n$ .
Any smaller proposed $B_n$ would fail for this single function and input.
The example also shows why a bound based only on adding local errors eventually stops being sharp: the finite range limits the total displacement.

Answer

$B_n=\min(n/5,1)$ . One function attaining every bound at $x=0$ is $f(x)=\min(x+1/5,1)$ .

Key idea

Combine accumulated local error with a global range restriction, then test sharpness using a deliberately constructed function.
Problem 3: Mark complete
Problem 4 When two rules can change order

Difficulty: 2 of 3 stars, Challenge

Define $F:\mathbb R\to\mathbb R$ by
$F(x)=\begin{cases}-2x&x\le0,\\-x/2&x>0.\end{cases}$
Let $G(x)=ax+b$ , where $a,b$ are real and the constant cases are allowed.

Find every pair $(a,b)$ for which $F(G(x))=G(F(x))$ for every real $x$ . Also prove that $F$ is its own inverse. Your classification must include functions $G$ that are not bijections.

Builds on Inverse Functions
1. Hint 1
  First evaluate the commutation equation at the unique fixed point of $F$ .
2. Hint 2
  After finding $b$ , separate $a>0$ , $a=0$ , and $a<0$ . Positive scaling preserves the branch of $F$ , while negative scaling exchanges branches.
Answer

Exactly $b=0$ and $a\ge0$ . Also $F^{-1}=F$ .

Full solution

The only fixed point of $F$ is $0$ : the equation $F(x)=x$ gives $-2x=x$ on the nonpositive branch and $-x/2=x$ on the positive branch.
Setting $x=0$ in the commutation equation gives $F(b)=b$ , so necessarily $b=0$ .

If $a>0$ , multiplying an input by $a$ preserves its sign, and each branch formula directly gives
$F(ax)=aF(x).$
Thus all positive $a$ work.
If $a=0$ , both compositions are the zero function, so this constant case works too.

If $a<0$ , use $x=1$ .
Then $F(G(1))=F(a)=-2a$ , whereas
$G(F(1))=a(-1/2)=-a/2.$
Equality would force $a=0$ , contradicting $a<0$ .
This excludes every negative setting.

Finally, if $x<0$ , the first application sends it to $-2x>0$ , and the second gives
$-(-2x)/2=x.$
If $x>0$ , the first sends it to $-x/2<0$ , and the second gives
$-2(-x/2)=x.$
Zero stays zero.
Thus $F(F(x))=x$ everywhere, proving $F$ is bijective with inverse $F$ .

Answer

Exactly $b=0$ and $a\ge0$ . Also $F^{-1}=F$ .

Key idea

Fixed points can constrain a commuting function before its full formulas are compared.
Problem 4: Mark complete
Problem 5 Every pair keeps its distance

Difficulty: 2 of 3 stars, Challenge

Find all functions $f:\mathbb R\to\mathbb R$ such that
$|f(x)-f(y)|=|x-y|\qquad\text{for all real }x,y.$
Do not assume that $f$ is linear, continuous, or monotone.

Your proof should explain why choosing the images of just two distinct inputs forces every other image, and should verify that every function in your answer works.
1. Hint 1
  Let $c=f(0)$ and use the distance between $0$ and $1$ to determine the two possibilities for $f(1)$ .
2. Hint 2
  For an arbitrary input $x$ , square its distance equations to $0$ and $1$ and subtract them. The unknown square of $f(x)-c$ will cancel.
Answer

Exactly $f(x)=x+c$ and $f(x)=-x+c$ , with arbitrary real $c$ .

Full solution

Put $c=f(0)$ .
The pair $0,1$ gives $|f(1)-c|=1$ , so write $f(1)=c+s$ , where $s=1$ or $s=-1$ .
For any real $x$ , let $u=f(x)-c$ .
Its distances to the images of $0$ and $1$ give
$u^2=x^2,\qquad(u-s)^2=(x-1)^2.$
Squaring is legitimate here because both original sides are nonnegative.

Subtracting the first squared equation from the second gives $-2su+1=-2x+1$ , hence $su=x$ .
Since $s^2=1$ , this forces $u=sx$ and therefore $f(x)=c+sx$ for every input.

Conversely, for either sign,
$|f(x)-f(y)|=|s(x-y)|=|x-y|.$
Thus both families work, with no additional conditions on $c$ .
The images of $0$ and $1$ chose the translation and the orientation; the two distance equations then fixed every remaining image.
Merely assigning $f(x)$ one of the two values $c\pm x$ independently at each input would fail: distances to a second anchor coordinate the sign choices.

Answer

Exactly $f(x)=x+c$ and $f(x)=-x+c$ , with arbitrary real $c$ .

Key idea

Distances to two anchors can force a global rule even when no regularity of the function is assumed.
Problem 5: Mark complete
Problem 6 How far apart can inverse readings lie?

Difficulty: 2 of 3 stars, Challenge

Let $f:[0,4]\to[0,8]$ be a strictly increasing bijection with $f(0)=0$ and $f(4)=8$ . For every $0\le x<y\le4$ , it satisfies
$1\le\frac{f(y)-f(x)}{y-x}\le3.$
Set $u=f^{-1}(3)$ and $v=f^{-1}(5)$ .

Find the exact set of possible values of $u+v$ as $f$ varies. Prove both bounds and construct a permitted function for every value between them. No differentiability is assumed.

Builds on Inverse Functions
1. Hint 1
  Apply the upper bound on the rate of change to the pairs $(0,u)$ , $(u,v)$ , and $(v,4)$ .
2. Hint 2
  For construction, use a polygonal graph through $(0,0),(u,3),(v,5),(4,8)$ , and choose $v-u=2/3$ . Check the slopes of its three pieces.
Answer

Exactly $8/3\le u+v\le16/3$ , with both endpoints attainable.

Full solution

Strict increase gives $0<u<v<4$ .
The rate bound applied to the three successive intervals gives $3/u\le3$ , $2/(v-u)\le3$ , and
$3/(4-v)\le3.$
Thus $u\ge1$ , $v-u\ge2/3$ , and $v\le3$ .
Consequently
$u+v=2u+(v-u)\ge8/3,$
while
$u+v=2v-(v-u)\le16/3.$

To realize every value in this interval, choose any $u\in[1,7/3]$ and set $v=u+2/3$ .
Join $(0,0),(u,3),(v,5),(4,8)$ by straight segments.
Their slopes are
$\frac3u,\qquad3,\qquad\frac3{10/3-u}.$
Both $u$ and $10/3-u$ lie in $[1,7/3]$ , so all slopes lie between $1$ and $3$ .

The resulting graph is continuous and strictly increasing from height $0$ to height $8$ , hence gives a bijection of the required intervals.
A secant crossing several pieces has slope equal to a length-weighted average of their slopes, so the required rate bound holds for every pair of inputs, not just within individual pieces.
Its inverse readings are exactly the chosen $u,v$ , and $u+v=2u+2/3$ fills $[8/3,16/3]$ .
This proves full attainability as well as sharp endpoint bounds.

Answer

Exactly $8/3\le u+v\le16/3$ , with both endpoints attainable.

Key idea

Bounds on inverse inputs require both necessary spacing inequalities and a function construction that satisfies every secant condition.
Problem 6: Mark complete
Problem 7 Two rules for a positive function

Difficulty: 2 of 3 stars, Challenge

Find all functions $f:(0,\infty)\to(0,\infty)$ satisfying
$f(xy)=f(x)f(y)\qquad(x,y>0)$
and
$f(t)+f(1-t)=1\qquad(0<t<1).$
Prove your answer without assuming continuity or a particular formula for $f$ . You may use the elementary fact that between any two distinct real numbers there is a rational number.
1. Hint 1
  Use $t=x/(x+y)$ in the second rule, and multiply by $f(x+y)$ .
2. Hint 2
  After obtaining $f(x+y)=f(x)+f(y)$ for positive inputs, positivity forces strict increase. Determine rational values and trap each real value between them.
Answer

The unique function is $f(x)=x$ for every $x>0$ .

Full solution

Multiplicativity gives $f(1)=f(1)^2$ , and positivity forces $f(1)=1$ .
For positive $x,y$ , put $t=x/(x+y)$ .
Multiplying $f(t)+f(1-t)=1$ by $f(x+y)$ and using multiplicativity gives
$f(x)+f(y)=f(x+y).$
Thus the second condition converts the multiplicative rule into an additive rule as well.

If $y>x>0$ , additivity gives $f(y)-f(x)=f(y-x)>0$ , so $f$ is strictly increasing.
Repeated addition gives $f(m)=m$ for positive integers $m$ .
Also $n f(m/n)=f(m)=m$ , so $f(q)=q$ for every positive rational $q$ .

Fix any real $x>0$ .
If $f(x)>x$ , choose a rational $q$ strictly between $x$ and $f(x)$ .
Strict increase would give $f(q)>f(x)$ , but $f(q)=q<f(x)$ , a contradiction.
If $f(x)<x$ , choose a positive rational $q$ strictly between $f(x)$ and $x$ ; then strict increase gives $q=f(q)<f(x)$ , again a contradiction.
Therefore $f(x)=x$ .
This function is positive and plainly satisfies both original rules, completing existence and uniqueness.

Answer

The unique function is $f(x)=x$ for every $x>0$ .

Key idea

Positivity can supply the order information needed to determine a function from its rational values.
Problem 7: Mark complete
Problem 8 How many inputs survive two stages?

Difficulty: 3 of 3 stars, Deep challenge

The graph of $f:[0,3]\to[0,3]$ consists exactly of the straight segments joining $(0,0),(1,3),(2,0),(3,2)$ , in that order, with all endpoints included.

(a) For each real $y$ , determine the number of distinct solutions of $f(f(x))=y$ with $x\in[0,3]$ . Account for the special levels $0$ , $2$ , and $3$ .

(b) Find all inputs for which $f(f(x))=1$ . Explain how counting preimages in two stages avoids writing a long piecewise formula for the composition.

Text description of this figure

Coordinate axes labeled x and y, each marked at 1, 2 and 3, with the origin labeled 0. The graph is a broken line of three straight segments with a dot at each of its four endpoints. It starts at the origin, rises to the point with x equal to 1 and y equal to 3, falls to the point on the x axis where x equals 2, and rises again to its end point, where x equals 3 and y equals 2. The peak at x equal to 1 and the end point at x equal to 3 are labeled with their coordinates.

Builds on Solving Linear Equations
1. Hint 1
  First count the preimages of a single level under $f$ . A vertex shared by two segments counts only once.
2. Hint 2
  For $0<y<2$ , the intermediate inputs are $y/3$ , $2-y/3$ , and $2+y/2$ . Count the preimages of each of those numbers separately.
Answer

(a) The count is $0$ outside $[0,3]$ ; $5$ at $0$ ; $8$ for $0<y<2$ ; $7$ at $2$ ; $6$ for $2<y<3$ ; and $3$ at $3$ . (b) $x=1/9,5/9,5/6,7/6,13/9,17/9,13/6,17/6$ .

Full solution

The three branch formulas are $3x$ on $[0,1]$ , $6-3x$ on $[1,2]$ , and $2x-4$ on $[2,3]$ .
Let $N(t)$ count the distinct preimages of $t$ under $f$ .
The graph gives $N(0)=2$ , $N(t)=3$ for $0<t\le2$ , $N(t)=2$ for $2<t<3$ , and $N(3)=1$ ; outside $[0,3]$ it is zero.

For $0<y<2$ , the three intermediate values satisfying $f(t)=y$ are $y/3$ , $2-y/3$ , and $2+y/2$ .
Their preimage counts are $3,3,2$ , totaling $8$ .
At $y=2$ , the last intermediate value is $3$ , with only one preimage, so the count is $7$ .
For $2<y<3$ , only the first two intermediate values remain, each with three preimages, giving $6$ .
At $y=0$ , the intermediate values are $0,2$ , giving $2+3=5$ ; at $y=3$ , the sole intermediate value is $1$ , giving $3$ .
No input can be counted for two distinct intermediate values because $f(x)$ has only one value.

For $y=1$ , the intermediate values are $1/3,5/3,5/2$ .
Their preimages are, respectively, $\{1/9,17/9,13/6\}$ , $\{5/9,13/9,17/6\}$ , and $\{5/6,7/6\}$ .
Combining and sorting gives the eight stated inputs.
Every branch used lies in its declared domain, so the two-stage enumeration is complete.

Answer

(a) The count is $0$ outside $[0,3]$ ; $5$ at $0$ ; $8$ for $0<y<2$ ; $7$ at $2$ ; $6$ for $2<y<3$ ; and $3$ at $3$ . (b) $x=1/9,5/9,5/6,7/6,13/9,17/9,13/6,17/6$ .

Key idea

Count a composition’s preimages by summing counts over distinct intermediate values; handle vertex levels separately.
Problem 8: Mark complete
Problem 9 A graph that settles after one application

Difficulty: 3 of 3 stars, Deep challenge

For real parameters $a,c$ , define a continuous piecewise linear function on $\mathbb R$ by
$f(x)=\begin{cases}ax&x<0,\\x&0\le x\le1,\\1+c(x-1)&x>1.\end{cases}$
Find all pairs $(a,c)$ for which $f(f(x))=f(x)$ for every real $x$ . For each family in your answer, describe the image of $f$ . Prove completeness, including zero slopes and negative slopes.
1. Hint 1
  The equation means that every output of $f$ must be a fixed point of $f$ .
2. Hint 2
  If $a>0$ , the left branch outputs negative values, which must themselves be fixed. If $a<0$ , that branch outputs every positive value. Make the corresponding observations for the right branch.
Answer

Exactly $(1,1)$ ; $(a,1)$ with $a\le0$ ; $(1,c)$ with $c\le0$ ; and $(0,0)$ . Their images are respectively $\mathbb R$ , $[0,\infty)$ , $(-\infty,1]$ , and $[0,1]$ .

Full solution

The required equation holds exactly when every value in the image is fixed by $f$ .
All of $[0,1]$ is already fixed.

If $a>0$ , negative inputs produce every negative output.
Every negative $t$ must then satisfy $at=t$ , forcing $a=1$ .
If $a<0$ , the left branch produces every positive output, including all $t>1$ ; these can all be fixed only if $c=1$ .
If $a=0$ , the left branch contributes only the already fixed value $0$ .

Similarly, if $c>0$ , the right branch produces every output above $1$ , forcing $c=1$ .
If $c<0$ , it produces every value below $1$ , including every negative value, forcing $a=1$ .
If $c=0$ , it contributes only the fixed value $1$ .
Combining these implications leaves precisely the four listed possibilities: both identity slopes; a nonpositive left slope with identity right slope; identity left slope with a nonpositive right slope; or two zero slopes.

Conversely, their respective images are $\mathbb R$ , $[0,\infty)$ , $(-\infty,1]$ , and $[0,1]$ .
On each listed image the function is the identity.
Thus every output is fixed, proving the composition equation for every case.
This also proves the image descriptions and includes the endpoint cases $a=0$ or $c=0$ without treating them as limits.

Answer

Exactly $(1,1)$ ; $(a,1)$ with $a\le0$ ; $(1,c)$ with $c\le0$ ; and $(0,0)$ . Their images are respectively $\mathbb R$ , $[0,\infty)$ , $(-\infty,1]$ , and $[0,1]$ .

Key idea

An equation of the form $f\circ f=f$ says that the image lies entirely inside the fixed-point set.
Problem 9: Mark complete
Problem 10 An algebraic rule with a global sign condition

Difficulty: 3 of 3 stars, Deep challenge

Find every function $f:\mathbb R\to\mathbb R$ such that
$f(x+y)=f(x)+f(y)+2xy\qquad\text{for all real }x,y,$
and $f(x)\ge0$ for every real $x$ .

No continuity, monotonicity, or polynomial formula is assumed. You may use the fact that between any two distinct real numbers there is a rational number. Explain where the nonnegativity condition rules out freedom that the algebraic identity alone would allow.
1. Hint 1
  Subtract $x^2$ : the function $g(x)=f(x)-x^2$ obeys $g(x+y)=g(x)+g(y)$ . First determine $g$ on rational inputs.
2. Hint 2
  If $g(1)=c\ne0$ , choose a rational $q$ between $0$ and $-c$ to contradict $q^2+cq\ge0$ . Then use rational shifts of an arbitrary real input.
Answer

The unique function is $f(x)=x^2$ for every real $x$ .

Full solution

Putting $x=y=0$ gives $f(0)=0$ .
Define
$g(x)=f(x)-x^2.$
Expanding $(x+y)^2$ shows $g(x+y)=g(x)+g(y)$ , so
$g(-x)=-g(x).$
If $c=g(1)$ , repeated addition and division give $g(q)=cq$ for every rational $q$ , including negative ones.

Nonnegativity at rational inputs says
$q(q+c)\ge0.$
If $c\ne0$ , a rational $q$ strictly between $0$ and $-c$ would make these two factors have opposite signs.
Therefore $c=0$ and $g(q)=0$ for every rational $q$ .

Fix an arbitrary real $x$ .
For every rational $q$ , additivity gives
$0\le f(x+q)=(x+q)^2+g(x).$
If $g(x)<0$ , choose a rational $q$ in the nonempty interval $(-x-\sqrt{-g(x)},-x+\sqrt{-g(x)})$ .
Then $(x+q)^2<-g(x)$ , a contradiction.
Hence $g(x)\ge0$ for every $x$ .
Applying this to $-x$ and using oddness gives $g(x)\le0$ as well.
Thus $g=0$ and $f(x)=x^2$ .

This function satisfies both original requirements.
The identity alone allows at least the functions $x^2+cx$ for arbitrary real $c$ , so the global sign condition is essential.
The proof uses it first to eliminate rational drift and then to eliminate any remaining drift at irrational inputs, without assuming regularity.

Answer

The unique function is $f(x)=x^2$ for every real $x$ .

Key idea

A global inequality can force a functional rule beyond its rational inputs, even without continuity.
Problem 10: Mark complete