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Functions: 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.

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Problem 1 of 10

Problem 1 Two names for one input

Difficulty: 1 of 3 stars, Stretch

A student proposes a function $f:\mathbb R\to\mathbb R$ satisfying $f(x^2-2x)=x+1$ for every real $x$ .

(a) Prove that no such function exists.

(b) Find all real constants $a,b$ for which a function $f:\mathbb R\to\mathbb R$ can satisfy $f(x^2-2x)=ax^2+bx+1$ for every real $x$ . Describe every possible $f$ , including any freedom in its definition.

Builds on What Is a Function?, Completing the Square
1. Hint 1
  Two different choices of $x$ can send the same number into $f$ .
2. Hint 2
  Compare $x=0$ with $x=2$ , then use the full range of $x^2-2x=(x-1)^2-1$ .
Answer

(a) Impossible. (b) Exactly $b=-2a$ , with arbitrary real $a$ ; then $f(t)=at+1$ for $t\ge-1$ , and values for $t<-1$ are arbitrary.

Full solution

In part (a), $x=0$ gives $f(0)=1$ , whereas $x=2$ gives $f(0)=3$ .
One input cannot have two outputs, so the proposed function cannot exist.

For part (b), the same two substitutions require $1=4a+2b+1$ , hence $b=-2a$ .
This is necessary even without assuming a formula for $f$ .
With this condition the right side becomes $a(x^2-2x)+1$ .

The possible inputs $t=x^2-2x=(x-1)^2-1$ are exactly all real numbers $t\ge-1$ : each is obtained by taking
$x=1+\sqrt{t+1}.$
Therefore the condition forces $f(t)=at+1$ on that entire interval.
Conversely, this rule gives the desired identity for every real $x$ whenever $b=-2a$ .

No number less than $-1$ ever appears as the input $x^2-2x$ .
The identity consequently puts no restriction on $f$ there.
Assigning any real value separately at each such input completes a valid function.
Thus the displayed formula need not hold on all of $\mathbb R$ .

Answer

(a) Impossible. (b) Exactly $b=-2a$ , with arbitrary real $a$ ; then $f(t)=at+1$ for $t\ge-1$ , and values for $t<-1$ are arbitrary.

Key idea

Before solving a functional equation, check whether different substitutions name the same input and which inputs are ever reached.
Problem 1: Mark complete
Problem 2 The missing inputs remain missing

Difficulty: 1 of 3 stars, Stretch

Let $f(x)=1/(x-1)$ on its natural real domain, and let $g(x)=(x+1)/x$ on its natural real domain.

Find the exact domain and range of each function and the exact domain of each composition $f\circ g$ and $g\circ f$ . Are $f$ and $g$ inverse functions? Explain why simplifying both compositions to $x$ does not make either composition defined for every real input.

Builds on Algebraic Fractions
1. Hint 1
  A composition needs both a valid first input and a valid intermediate output.
2. Hint 2
  To establish a range, solve $y=f(x)$ or $y=g(x)$ for $x$ and determine exactly which $y$ values are possible.
Answer

$f:\mathbb R\setminus\{1\}\to\mathbb R\setminus\{0\}$ and $g:\mathbb R\setminus\{0\}\to\mathbb R\setminus\{1\}$ are inverses. The composition domains are respectively $\mathbb R\setminus\{0\}$ and $\mathbb R\setminus\{1\}$ .

Full solution

The formula for $f$ requires $x\ne1$ , and its output cannot be $0$ .
Every nonzero real $y$ is obtained at $x=1+1/y$ , which is allowed.
Thus its range is exactly $\mathbb R\setminus\{0\}$ .
Similarly, $g(x)=1+1/x$ requires $x\ne0$ and never gives $1$ .
Every $y\ne1$ is obtained at $x=1/(y-1)$ , so its range is exactly $\mathbb R\setminus\{1\}$ .

For $f(g(x))$ , the inner function requires $x\ne0$ .
Its output never equals $1$ , so the outer function adds no restriction.
On this domain,
$f(g(x))=1/[(x+1)/x-1]=x.$

For $g(f(x))$ , the inner restriction is $x\ne1$ , and its nonzero output always lies in the domain of $g$ .
On this domain,
$g(f(x))=1+1/[1/(x-1)]=x.$

Both compositions are the appropriate identity functions on the original domains, so these are inverse functions.
Cancellation has simplified their values where they exist; it cannot supply the undefined first step at a missing input.

Answer

$f:\mathbb R\setminus\{1\}\to\mathbb R\setminus\{0\}$ and $g:\mathbb R\setminus\{0\}\to\mathbb R\setminus\{1\}$ are inverses. The composition domains are respectively $\mathbb R\setminus\{0\}$ and $\mathbb R\setminus\{1\}$ .

Key idea

A function includes its domain; an algebraically simplified formula alone does not specify the same function.
Problem 2: Mark complete
Problem 3 The largest reversible interval

Difficulty: 1 of 3 stars, Stretch

Consider $f(x)=x^2-6x+5$ . Find the largest interval containing $5$ on which the restriction of $f$ has an inverse function. Include or exclude boundary points as appropriate, prove that no larger interval containing $5$ works, and give the inverse with its domain and range.

Builds on Completing the Square
1. Hint 1
  The same output occurs at inputs reflected across the vertex.
2. Hint 2
  Write $f(x)=(x-3)^2-4$ . The inverse must choose a root consistent with the chosen interval.
Answer

The interval is $[3,\infty)$ . Its inverse is $f^{-1}(y)=3+\sqrt{y+4}$ , with domain $[-4,\infty)$ and range $[3,\infty)$ .

Full solution

Completing the square gives
$f(x)=(x-3)^2-4.$
On $[3,\infty)$ , increasing the input strictly increases the nonnegative number $x-3$ , and hence its square.
The restriction is one-to-one and contains the input $5$ .
Its range is $[-4,\infty)$ , since the square takes every nonnegative value.

Any interval containing $5$ and any point $u<3$ must also contain two distinct points $3-\varepsilon$ and $3+\varepsilon$ for a sufficiently small positive $\varepsilon$ .
Those points give the same output $\varepsilon^2-4$ , so such an interval is not one-to-one.
Therefore every interval containing $5$ that works is contained in $[3,\infty)$ , establishing the claimed largest interval.

For a given output $y\ge-4$ , the equation $(x-3)^2=y+4$ has two formal root choices, but $x\ge3$ selects
$x=3+\sqrt{y+4}.$
This gives exactly one allowed input for every allowed output.
Its domain is $[-4,\infty)$ and its range is $[3,\infty)$ , the reverse of those of the restricted function.

Answer

The interval is $[3,\infty)$ . Its inverse is $f^{-1}(y)=3+\sqrt{y+4}$ , with domain $[-4,\infty)$ and range $[3,\infty)$ .

Key idea

Choosing an inverse branch is a domain decision, and maximality must be justified using repeated outputs.
Problem 3: Mark complete
Problem 4 One-way undoing on a finite set

Difficulty: 2 of 3 stars, Challenge

Let $S=\{-4,-3,-2,-1,0,1,2,3,4\}$ and $T=\{0,1,4,9,16\}$ . Define $f:S\to T$ by $f(x)=x^2$ .

(a) How many functions $g:T\to S$ satisfy $f(g(t))=t$ for every $t\in T$ ?

(b) Find all such $g$ for which the sum of their five output values is $0$ .

(c) Can any function $h:T\to S$ satisfy $h(f(x))=x$ for every $x\in S$ ? Prove your answer.

Builds on What Is a Function?
1. Hint 1
  For each square in $T$ , choose one input that maps to it under $f$ .
2. Hint 2
  For part (b), choosing signs for $1,2,3,4$ gives sum zero exactly when the positive magnitudes sum to half of $1+2+3+4$ .
Answer

(a) $16$ . (b) In input order $(0,1,4,9,16)$ , the outputs are $(0,1,-2,-3,4)$ or $(0,-1,2,3,-4)$ . (c) No.

Full solution

The condition forces $g(0)=0$ .
For each of $1,4,9,16$ , the output of $g$ can independently be either square root in $S$ : respectively $\pm1,\pm2,\pm3,\pm4$ .
Each choice works because squaring recovers the input.
Thus there are $2^4=16$ functions.

For part (b), the magnitudes sum to $10$ .
The signed sum is zero precisely when the magnitudes assigned positive signs sum to $5$ .
A single magnitude cannot do this, and three magnitudes already sum to at least $1+2+3=6$ .
The only two-element subsets summing to $5$ are $\{1,4\}$ and $\{2,3\}$ .
These give exactly the two output lists stated.

For part (c),
$f(1)=f(-1)=1.$
The proposed condition would require both $h(1)=1$ and $h(1)=-1$ , which is impossible.
Although each of the $16$ functions in part (a) chooses a valid way back from an output, none can recover both original inputs when the same square has two sources.
A one-way undoing condition does not guarantee an inverse.

Answer

(a) $16$ . (b) In input order $(0,1,4,9,16)$ , the outputs are $(0,1,-2,-3,4)$ or $(0,-1,2,3,-4)$ . (c) No.

Key idea

Choosing one preimage of every output can undo a function in one direction even when information has been lost.
Problem 4: Mark complete
Problem 5 Returning after two steps

Difficulty: 2 of 3 stars, Challenge

Let $f(x)=x^2-2$ for all real $x$ . Find every real $x$ such that $f(f(x))=x$ . Separate the inputs that return after one step from those that return only after two steps, and identify the two-step pairs. Prove your list is complete without expanding a quartic.
1. Hint 1
  Name the intermediate output $y=f(x)$ . Then the condition says $x=f(y)$ .
2. Hint 2
  Subtract $y=x^2-2$ and $x=y^2-2$ . One factor distinguishes a fixed point from a two-step pair.
Answer

Fixed inputs: $-1,2$ . The remaining inputs are $(-1+\sqrt5)/2$ and $(-1-\sqrt5)/2$ , which map to each other.

Full solution

Put $y=f(x)$ .
The condition is equivalent to the pair $y=x^2-2$ and $x=y^2-2$ .
Subtracting gives $y-x=x^2-y^2=(x-y)(x+y)$ , hence
$(x-y)(x+y+1)=0.$
Every solution therefore belongs to one of two cases.

If $x=y$ , then $x=x^2-2$ , so
$(x-2)(x+1)=0.$
The inputs $2$ and $-1$ are fixed points and satisfy the original condition.

If $x\ne y$ , then $x+y=-1$ .
Substituting $y=-1-x$ into $y=x^2-2$ yields $x^2+x-1=0$ , whose roots are $\alpha=(-1+\sqrt5)/2$ and
$\beta=(-1-\sqrt5)/2.$
Each satisfies $x^2-2=-1-x$ , and $\alpha+\beta=-1$ , so $f(\alpha)=\beta$ and
$f(\beta)=\alpha.$
They are distinct and neither is a fixed point.

The two cases exhaust the factorized equation, and every candidate has been checked in the original two-step process.
Thus these four inputs are the complete list.

Answer

Fixed inputs: $-1,2$ . The remaining inputs are $(-1+\sqrt5)/2$ and $(-1-\sqrt5)/2$ , which map to each other.

Key idea

For a two-step return, introduce the intermediate state and use the symmetry between the two equations.
Problem 5: Mark complete
Problem 6 Which invented operations associate?

Difficulty: 2 of 3 stars, Challenge

Fix real constants $p,q,r$ and define an operation on all real numbers by $a\star b=pa+qb+r$ .

(a) Find every triple $(p,q,r)$ for which $(a\star b)\star c=a\star(b\star c)$ for all real $a,b,c$ .

(b) Among these operations, identify every one possessing a two-sided identity: a real $e$ with $a\star e=e\star a=a$ for all real $a$ . Give the identity in each case.
1. Hint 1
  Associativity is an identity in three independent inputs, not an equation at a few test values.
2. Hint 2
  Expand both parenthesizations and compare the coefficients of $a$ and $c$ , then compare constants.
Answer

(a) $(p,q)=(0,0)$ or $(1,1)$ with arbitrary $r$ ; or $(p,q,r)=(1,0,0),(0,1,0)$ . (b) Exactly $a\star b=a+b+r$ , with identity $e=-r$ .

Full solution

Expanding the left parenthesization gives $p^2a+pqb+qc+pr+r$ ; expanding the right gives $pa+pqb+q^2c+qr+r$ .
Their difference must vanish for every independent choice of $a,b,c$ .
Therefore

$p^2=p,\qquad q^2=q,\qquad (p-q)r=0.$

The first two conditions force $p,q\in\{0,1\}$ .
If they are equal, the third condition allows any $r$ , giving the constant operation $a\star b=r$ and the shifted sum
$a\star b=a+b+r.$
If they differ, $r=0$ , giving the projections $a\star b=a$ and
$a\star b=b.$
The expansion also proves each listed family is associative, so the classification is complete.

A constant operation cannot have an identity for every real $a$ .
For the first projection, $e\star a=e$ fails to equal arbitrary $a$ ; the second projection fails on the other side.
Finally, for the shifted sum, $a\star e=a+e+r=a$ and $e\star a=e+a+r=a$ hold exactly when $e=-r$ .
This identity is unique.

Answer

(a) $(p,q)=(0,0)$ or $(1,1)$ with arbitrary $r$ ; or $(p,q,r)=(1,0,0),(0,1,0)$ . (b) Exactly $a\star b=a+b+r$ , with identity $e=-r$ .

Key idea

Universal operation laws become coefficient conditions, but an identity must work on both sides.
Problem 6: Mark complete
Problem 7 A fraction operation with hidden multiplication

Difficulty: 2 of 3 stars, Challenge

For real numbers $a,b$ with $-1<a,b<1$ , define $a\diamond b=(a+b)/(1+ab)$ .

(a) Prove that the output is again between $-1$ and $1$ , and prove that the operation is associative.

(b) Find all $x\in(-1,1)$ such that $(x\diamond\tfrac12)\diamond\tfrac13=\tfrac34$ .

An associativity proof must apply to all permitted inputs, not only the numbers in part (b).

Builds on Algebraic Fractions
1. Hint 1
  First inspect $1-(a\diamond b)$ and $1+(a\diamond b)$ .
2. Hint 2
  Try the new coordinate $\phi(t)=(1+t)/(1-t)$ . What is $\phi(a\diamond b)$ in terms of $\phi(a)$ and $\phi(b)$ ?
Answer

(a) The operation is closed on $(-1,1)$ and associative. (b) The unique solution is $x=1/13$ .

Full solution

Since $ab>-1$ , the denominator $1+ab$ is positive.
Also $1-(a\diamond b)=(1-a)(1-b)/(1+ab)>0$ , and
$1+(a\diamond b)=(1+a)(1+b)/(1+ab)>0.$
These prove $-1<a\diamond b<1$ , so repeated operations are defined.

Define
$\phi(t)=(1+t)/(1-t).$
This sends $(-1,1)$ one-to-one onto the positive real numbers: solving $z=\phi(t)$ gives $t=(z-1)/(z+1)$ for every $z>0$ .
The two factorizations above give

$\phi(a\diamond b)=\frac{(1+a)(1+b)}{(1-a)(1-b)}=\phi(a)\phi(b).$

Therefore both $(a\diamond b)\diamond c$ and $a\diamond(b\diamond c)$ have the same $\phi$ -value, namely $\phi(a)\phi(b)\phi(c)$ .
Since $\phi$ is one-to-one, they are equal.
This proves associativity for every allowed triple.

For part (b), the equation becomes
$\phi(x)\cdot3\cdot2=7.$
Thus $\phi(x)=7/6$ , and
$x=[(7/6)-1]/[(7/6)+1]=1/13.$
This lies in the domain.
Every step was reversible, so it is the unique solution.

Answer

(a) The operation is closed on $(-1,1)$ and associative. (b) The unique solution is $x=1/13$ .

Key idea

A change of coordinates can turn an unfamiliar operation into an ordinary one and transfer its laws.
Problem 7: Mark complete
Problem 8 Can a finite function have a square root?

Difficulty: 3 of 3 stars, Deep challenge

Let $S=\{1,2,3,4,5\}$ . A function $f:S\to S$ is called a square root of $h:S\to S$ here if $f(f(n))=h(n)$ for every $n\in S$ .

(a) Does a square root exist when the outputs of $h$ , in input order $1,2,3,4,5$ , are $(2,1,3,4,5)$ ?

(b) Find every square root when the outputs of $h$ are $(2,1,4,3,5)$ . Prove completeness in both cases. You do not need any theorems about permutations.
1. Hint 1
  Because each proposed $h$ takes distinct inputs to distinct outputs, the same must be true of $f$ .
2. Hint 2
  Show $f(h(n))=h(f(n))$ . Thus $f$ must send inputs fixed by $h$ to inputs fixed by $h$ . In part (b), begin with the possibilities for $f(1)$ .
Answer

(a) None. (b) Exactly two: output lists $(3,4,2,1,5)$ and $(4,3,1,2,5)$ .

Full solution

In either case, $h$ is one-to-one.
If $f(u)=f(v)$ , applying $f$ gives $h(u)=h(v)$ and hence $u=v$ .
Thus $f$ is one-to-one, and on the finite set $S$ it uses every output exactly once.
Furthermore
$f(h(n))=f(f(f(n)))=h(f(n)).$
Consequently $f$ sends the set of fixed inputs of $h$ into itself, and, by its one-to-one property, onto that same set.

In part (a), the fixed set is $\{3,4,5\}$ , so $f$ must also map $\{1,2\}$ onto itself.
On two inputs it either leaves both in place or swaps them.
Doing either operation twice leaves both in place, whereas $h$ swaps them.
This is impossible.

In part (b), only $5$ is fixed, so $f(5)=5$ .
The value $f(1)$ cannot be $1$ , since $f(f(1))=2$ .
It cannot be $2$ , since that would force $f(2)=2$ and violate one-to-one behavior.
It is not $5$ .
Thus it is $3$ or $4$ .

If $f(1)=3$ , then $f(3)=2$ , then $f(2)=4$ , then $f(4)=1$ , using $f^2=h$ successively.
If $f(1)=4$ , the same reasoning gives $f(4)=2$ , $f(2)=3$ , $f(3)=1$ .
Directly composing each output list with itself gives the required $h$ , proving existence and completeness.

Answer

(a) None. (b) Exactly two: output lists $(3,4,2,1,5)$ and $(4,3,1,2,5)$ .

Key idea

Iteration constrains how a function moves whole cycles and fixed sets, not only individual values.
Problem 8: Mark complete
Problem 9 Three substitutions close the loop

Difficulty: 3 of 3 stars, Deep challenge

Let $D=\mathbb R\setminus\{0,1\}$ and define $T(x)=1/(1-x)$ for $x\in D$ . Find every function $f:D\to\mathbb R$ satisfying
$f(x)+f(T(x))=x\qquad\text{for every }x\in D.$
Your solution must justify that all compositions used stay in $D$ and must prove both existence and uniqueness. Then evaluate $f(2)$ .

Builds on Algebraic Fractions
1. Hint 1
  Compute $T(T(x))$ and $T(T(T(x)))$ before guessing a formula for $f$ .
2. Hint 2
  Write the given equation at the three inputs in this loop. Combine the three equations to isolate $f(x)$ .
Answer

The unique function is $f(x)=\tfrac12\left(x-\frac1{1-x}+\frac{x-1}{x}\right)$ on $D$ , and $f(2)=7/4$ .

Full solution

For $x\in D$ , $T(x)$ is defined and nonzero.
It equals $1$ only if $x=0$ , which is excluded.
Thus $T$ maps $D$ into itself.
Direct calculation gives $T^2(x)=(x-1)/x$ and $T^3(x)=x$ , with all denominators valid on $D$ .
Here $T^2$ and $T^3$ mean repeated composition.

Write $u=T(x)$ and $v=T^2(x)$ .
Applying the identity at $x,u,v$ gives $f(x)+f(u)=x$ , $f(u)+f(v)=u$ , and
$f(v)+f(x)=v.$
Add the first and third equations and subtract the second to obtain
$2f(x)=x-u+v.$
This forces the stated formula at every allowed input, so there can be at most one solution.

To check existence, define
$f(x)=(x-T(x)+T^2(x))/2.$
Then $f(T(x))=(T(x)-T^2(x)+x)/2$ because $T^3(x)=x$ .
Adding these gives $x$ , exactly as required.
Thus the formula does define a solution on the whole domain.

Finally, $T(2)=-1$ and $T^2(2)=1/2$ , so
$f(2)=(2+1+1/2)/2=7/4.$
No continuity or polynomial assumption was needed.

Answer

The unique function is $f(x)=\tfrac12\left(x-\frac1{1-x}+\frac{x-1}{x}\right)$ on $D$ , and $f(2)=7/4$ .

Key idea

When substitutions return to their starting input, a functional equation can become a finite linear system.
Problem 9: Mark complete
Problem 10 A square root of a shift

Difficulty: 3 of 3 stars, Deep challenge

Find every function $f:\mathbb Z\to\mathbb Z$ such that $f(f(n))=n+2$ for every integer $n$ . Give a complete description that constructs each function, prove that every function described works, and exhibit one that is not $f(n)=n+1$ .
1. Hint 1
  Compute $f(f(f(n)))$ in two ways to relate $f(n+2)$ and $f(n)$ .
2. Hint 2
  Once $f(0)$ and $f(1)$ are known, the relation determines every even and odd input. Examine whether $f(0)$ could be even.
Answer

For every integer $t$ there is one solution: $f(2k)=2k+2t+1$ and $f(2k+1)=2k+2-2t$ for all integers $k$ . These are all solutions; $t=1$ gives a nontranslation example.

Full solution

Associating the threefold composition in two ways gives
$f(n+2)=f(f(f(n)))=f(n)+2.$
Repeatedly applying this relation, forwards and backwards, shows that if $a=f(0)$ and $b=f(1)$ , then $f(2k)=a+2k$ and $f(2k+1)=b+2k$ for every integer $k$ .

If $a$ were even, the even-input formula would give
$f(a)=a+a=2a.$
But the original condition at $n=0$ requires $f(a)=2$ , forcing $a=1$ , contrary to its being even.
Hence $a=2t+1$ for some integer $t$ .
The odd-input formula now gives $f(a)=b+2t=2$ , so $b=2-2t$ .
These are exactly the formulas in the answer, and every solution has been forced into this family.

Conversely, choose any integer $t$ and use those formulas to define $f$ .
An even input $2k$ maps to $2(k+t)+1$ , which then maps to
$2(k+t)+2-2t=2k+2.$
An odd input $2k+1$ maps to $2(k+1-t)$ , which then maps to
$2(k+1-t)+2t+1=2k+3.$
Thus $f(f(n))=n+2$ for both parities and all integers.

For $t=1$ , even inputs map to $n+3$ and odd inputs to $n-1$ .
This function satisfies the condition but is not the uniform shift $n+1$ .

Answer

For every integer $t$ there is one solution: $f(2k)=2k+2t+1$ and $f(2k+1)=2k+2-2t$ for all integers $k$ . These are all solutions; $t=1$ gives a nontranslation example.

Key idea

Commuting an unknown function with its own iterate can reveal a recurrence that reduces an infinite problem to a few values.
Problem 10: Mark complete