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Rational Expressions and 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 ways to group division

Difficulty: 1 of 3 stars, Stretch

Let $A,B,C$ be real numbers with $B\ne0$ and $C\ne0$ . Determine exactly when $(A/B)/C=A/(B/C)$ , including the possibility $A=0$ .

Now take $A=x(x-1)$ , $B=x+3$ , and $C=(x-2)/(x+1)$ . Find the original real domain of both grouped expressions and every input at which they are equal. Explain why dividing both sides by $A$ would give an incomplete answer.
1. Hint 1
  Express the left side as $A/(BC)$ and the right side as $AC/B$ , then clear only known nonzero factors.
2. Hint 2
  The general equality reduces to $A(1-C^2)=0$ . In the application, handle $A=0$ before solving $C=1$ and $C=-1$ .
Answer

The general condition is $A=0$ or $C=\pm1$ . The common original domain is $\mathbb R\setminus\{-3,-1,2\}$ , and equality occurs exactly at $x=0,1,1/2$ .

Full solution

Both grouped expressions are defined because $B$ and $C$ are nonzero.
They simplify to $A/(BC)$ and $AC/B$ , respectively.
Multiplying their equality by the nonzero $BC$ gives $A=AC^2$ , or
$A(1-C^2)=0.$
Therefore equality holds precisely when $A=0$ or $C=1$ or $C=-1$ .
Each of these conditions also directly makes the original expressions equal.

In the application, defining $C$ excludes $x=-1$ .
Both groupings require $B\ne0$ , excluding $x=-3$ , and $C\ne0$ , excluding $x=2$ .
No other exclusion arises: for the right grouping, the intermediate denominator $B/C$ is nonzero exactly when these conditions hold.

On this domain, $A=0$ gives $x=0,1$ , both permitted.
The equation $C=1$ would require $x-2=x+1$ , which is impossible.
The equation $C=-1$ gives $x-2=-x-1$ , hence $x=1/2$ , also permitted.
These exhaust the general criterion and so exhaust the answers.
Dividing the equality by $A$ at the start would improperly discard the valid inputs $0$ and $1$ .
Parentheses change division because dividing by a quotient multiplies by its reciprocal.

Answer

The general condition is $A=0$ or $C=\pm1$ . The common original domain is $\mathbb R\setminus\{-3,-1,2\}$ , and equality occurs exactly at $x=0,1,1/2$ .

Key idea

Before canceling a variable factor, separate its zero case; grouping division changes which factors become reciprocals.
Problem 1: Mark complete
Problem 2 Three fractions with one numerator

Difficulty: 1 of 3 stars, Stretch

Find all real solutions of
$\frac{x}{x+1}+\frac{x}{x+3}+\frac{x}{x+5}=0.$
Locate each solution relative to the forbidden inputs $-5,-3,-1$ , using exact comparisons rather than decimal approximations.
1. Hint 1
  Factor out the common numerator before multiplying all three denominators.
2. Hint 2
  Keep the possibility $x=0$ . For $x\ne0$ , the numerator of the reciprocal sum becomes especially simple after the shift $u=x+3$ .
Answer

$x=0$ and $x=-3\pm2\sqrt3/3$ . The latter zeros lie in $(-5,-3)$ and $(-3,-1)$ , respectively.

Full solution

The domain excludes
$x=-1,-3,-5.$
On that domain the equation is
$x(1/(x+1)+1/(x+3)+1/(x+5))=0.$
Thus $x=0$ is one valid solution.
Dividing immediately by $x$ would lose it.

For the other solutions, combine the three reciprocal fractions.
Their numerator is $(x+3)(x+5)+(x+1)(x+5)+(x+1)(x+3)=3x^2+18x+23=3(x+3)^2-4$ .
Since the common denominator is nonzero on the domain, this numerator must be zero.
Hence
$x=-3\pm2/\sqrt3=-3\pm2\sqrt3/3.$

The inequalities $0<2/\sqrt3<2$ follow from $\sqrt3>1$ .
Therefore the smaller root lies strictly between $-5$ and $-3$ , and the larger strictly between $-3$ and $-1$ .
Neither is forbidden, and the additional solution $0$ lies to the right of all three forbidden inputs.
Each manipulation was reversible after retaining the $x=0$ branch, which proves both validity and completeness.

Answer

$x=0$ and $x=-3\pm2\sqrt3/3$ . The latter zeros lie in $(-5,-3)$ and $(-3,-1)$ , respectively.

Key idea

Preserve a zero-numerator branch before reducing a rational equation to a common denominator.
Problem 2: Mark complete
Problem 3 Reconstructing a rational graph

Difficulty: 1 of 3 stars, Stretch

A rational expression $R(x)=P(x)/Q(x)$ has monic real polynomials $P,Q$ of degrees $3,2$ , respectively. Its original domain excludes exactly $-1$ and $2$ . At $x=-1$ its graph has a removable hole, and at $x=2$ it has a vertical asymptote. Its slant asymptote is $y=x+1$ , and $R(3)=0$ .

Determine $P$ and $Q$ . Find the height of the hole, and determine whether that height is attained anywhere else on the original graph. Justify uniqueness.
1. Hint 1
  The degree and domain determine $Q$ . Cancel the factor responsible for the hole, then use polynomial division and the slant asymptote.
2. Hint 2
  The reduced expression must have the form $x+1+c/(x-2)$ . Use the given zero to find $c$ , and then solve for the height of the hole using the original domain.
Answer

$Q(x)=(x+1)(x-2)$ , $P(x)=(x+1)(x-3)(x+2)$ . The hole is at $(-1,4/3)$ , and its height is also attained at $x=10/3$ .

Full solution

Because $Q$ is monic of degree $2$ with exactly the two real zeros $-1,2$ , it must equal $(x+1)(x-2)$ .
The removable hole forces the factor $x+1$ to cancel.
After this cancellation, division of the remaining monic quadratic by $x-2$ gives a linear quotient and a constant remainder.
The slant asymptote fixes the quotient as $x+1$ , so the reduced expression is $x+1+c/(x-2)$ .

The condition $R(3)=0$ gives $4+c=0$ , hence $c=-4$ .
Therefore the reduced numerator is $(x+1)(x-2)-4=x^2-x-6=(x-3)(x+2)$ , giving the claimed $P$ .
Its value at $2$ is nonzero, so the specified vertical asymptote really occurs.

At the excluded input $-1$ , the reduced formula has value $4/3$ .
To see whether this output is lost altogether, solve
$x+1-4/(x-2)=4/3.$
After multiplying by the nonzero $3(x-2)$ , the equation becomes
$3x^2-7x-10=(x+1)(3x-10)=0.$
The first root is excluded, but $10/3$ is permitted.
Thus the hole removes one point without removing its height from the range.
Each coefficient was forced in order, proving uniqueness.

Answer

$Q(x)=(x+1)(x-2)$ , $P(x)=(x+1)(x-3)(x+2)$ . The hole is at $(-1,4/3)$ , and its height is also attained at $x=10/3$ .

Key idea

A hole removes an input-output pair; it removes an output only if that output has no other preimage.
Problem 3: Mark complete
Problem 4 Returning after four steps

Difficulty: 2 of 3 stars, Challenge

Let $T(x)=(x-1)/(x+1)$ for real $x\ne-1$ . Starting from a real input $x$ , apply $T$ repeatedly, stopping if an input of $-1$ occurs.

Find exactly which starting values allow four applications. For every such starting value, determine $T^4(x)$ , where $T^4$ means four successive applications. Prove that the first return to the starting value occurs after exactly four applications.
1. Hint 1
  Compute the iterates one at a time. At each step, exclude inputs using the preceding expression before canceling factors.
2. Hint 2
  The second iterate simplifies to $-1/x$ , but it retains exclusions from the first step. Check returns after one, two, and three applications separately.
Answer

Four applications are possible exactly for $x\in\mathbb R\setminus\{-1,0,1\}$ . On this set, $T^4(x)=x$ , and the first return is always at step $4$ .

Full solution

The first application requires $x\ne-1$ .
For a second application, $T(x)$ must also differ from $-1$ ; solving $(x-1)/(x+1)=-1$ gives $x=0$ .
On the remaining domain,
$T^2(x)=-1/x.$
A third application fails exactly when $-1/x=-1$ , or $x=1$ .
Otherwise
$T^3(x)=-(x+1)/(x-1).$

For a fourth application, this last value must differ from $-1$ .
The equality $-(x+1)/(x-1)=-1$ would imply $-1=1$ , so it introduces no new exclusion.
Direct substitution now gives $T^4(x)=x$ for precisely the domain $\mathbb R\setminus\{-1,0,1\}$ .

A return after one application would satisfy $(x-1)/(x+1)=x$ , which reduces to $x^2=-1$ .
A return after two would satisfy $-1/x=x$ , again giving $x^2=-1$ .
A return after three would give $-(x+1)/(x-1)=x$ , with the same impossible real equation.
Thus no earlier return occurs.
In particular, the simplified formula $T^4(x)=x$ cannot be used to restore the three inputs at which the original four-step process breaks down.

Answer

Four applications are possible exactly for $x\in\mathbb R\setminus\{-1,0,1\}$ . On this set, $T^4(x)=x$ , and the first return is always at step $4$ .

Key idea

The domain of an iterate records every intermediate denominator, even if its final formula is simple.
Problem 4: Mark complete
Problem 5 Integer outputs of a rational function

Difficulty: 2 of 3 stars, Challenge

Find every integer $n$ for which
$F(n)=\frac{n^3+3n^2+14n+12}{n^2+2n+2}$
is an integer. Give all resulting integer outputs, and prove that no integer input is missing.
1. Hint 1
  Write $m=n+1$ and divide the numerator by the denominator.
2. Hint 2
  If $m^2+1$ divides $10m$ , use the identity $(m^2+1)-m\cdot m=1$ to show that $m^2+1$ divides $10$ .
Answer

The inputs are $n=-4,-3,-2,-1,0,1,2$ . The outputs are $-6$ for the first three, $0$ at $-1$ , and $6$ for the last three.

Full solution

The denominator is $(n+1)^2+1>0$ , so every real input is in the domain.
Let $m=n+1$ .
The numerator becomes $m^3+11m$ , and division gives
$F(n)=m+10m/(m^2+1).$
For integer $m$ , the output is an integer exactly when $m^2+1$ divides $10m$ .

Write $d=m^2+1$ .
If $d$ divides $10m$ , it divides $10m^2$ as well.
It also divides $10d$ , so it divides their difference
$10d-10m^2=10.$
Conversely, if $d$ divides $10$ , then it certainly divides $10m$ .
Thus the integrality condition is equivalent to $m^2+1$ being a positive divisor of $10$ .

These divisors are $1,2,5,10$ , giving $m^2=0,1,4,9$ and
$m=0,\pm1,\pm2,\pm3.$
Subtracting $1$ gives precisely the seven listed inputs.
At $m=1,2,3$ , the formula $m+10m/(m^2+1)$ gives $6$ each time.
It is odd as a function of $m$ , giving $-6$ at the three negative values, and $0$ at $m=0$ .
The divisibility equivalence supplies the required finite completeness argument.

Answer

The inputs are $n=-4,-3,-2,-1,0,1,2$ . The outputs are $-6$ for the first three, $0$ at $-1$ , and $6$ for the last three.

Key idea

Polynomial division can reduce integrality to divisibility, and a coprime denominator can sharply bound the search.
Problem 5: Mark complete
Problem 6 A rational expression on a triangle

Difficulty: 2 of 3 stars, Challenge

Positive real numbers $a,b,c$ satisfy $a+b+c=1$ . Prove the sharp bounds
$\frac34\le \frac{ab}{(1-a)(1-b)}+\frac{bc}{(1-b)(1-c)}+\frac{ca}{(1-c)(1-a)}<1.$
Identify every equality case for the lower bound. Prove that the upper bound cannot be replaced by any smaller number, even though it is never attained.
1. Hint 1
  Use a common denominator and abbreviate $q=ab+bc+ca$ and $p=abc$ .
2. Hint 2
  The expression simplifies to $(q-3p)/(q-p)$ . Prove $q/p=1/a+1/b+1/c\ge9$ by expanding $(a+b+c)(1/a+1/b+1/c)$ and pairing reciprocal ratios.
Answer

The bounds hold. Equality at $3/4$ occurs only when $a=b=c=1/3$ ; the upper bound $1$ is unattained but sharp.

Full solution

All three numbers lie in $(0,1)$ , so every denominator is positive.
With $q=ab+bc+ca$ and $p=abc>0$ , the common denominator is $(1-a)(1-b)(1-c)=q-p$ and the numerator is $q-3p$ .
Thus
$E=1-2p/(q-p)<1.$

To prove the lower bound, expand $(a+b+c)(1/a+1/b+1/c)=3+(a/b+b/a)+(b/c+c/b)+(c/a+a/c)$ .
Each paired sum is at least $2$ , since
$u/v+v/u-2=(u-v)^2/(uv)\ge0.$
Therefore
$q/p=1/a+1/b+1/c\ge9.$
It follows that
$E-3/4=(q-9p)/(4(q-p))\ge0.$
Equality requires equality in every paired ratio, so exactly $a=b=c=1/3$ works.

For sharpness above, take $a=b=(1-t)/2$ and $c=t$ , with $0<t<1$ .
Then $p=t(1-t)^2/4$ and $q-p=(1-t)(1+t)^2/4$ , yielding
$E=1-2t(1-t)/(1+t)^2.$
Its positive deficit below $1$ is smaller than $2t$ .
Given any $\varepsilon>0$ , choose $0<t<\min(1,\varepsilon/2)$ ; then
$E>1-\varepsilon.$
This proves sharpness without claiming an unattainable equality.

Answer

The bounds hold. Equality at $3/4$ occurs only when $a=b=c=1/3$ ; the upper bound $1$ is unattained but sharp.

Key idea

Distinguish an attained maximum from a sharp upper bound approached by valid inputs.
Problem 6: Mark complete
Problem 7 Lines meeting a rational curve

Difficulty: 2 of 3 stars, Challenge

For real $m,b$ , count the distinct real intersections of the line $y=mx+b$ with the curve $y=(x^2+1)/(x-1)$ , whose domain is $x\ne1$ . Give a complete classification in terms of $m,b$ , including every line with exactly one intersection. Explain the exceptional behavior when $m=1$ .
1. Hint 1
  Clear $x-1$ only after recording the excluded input, and check whether that input could become a root.
2. Hint 2
  When $m\ne1$ , the discriminant can be written as $(b+m-2)^2+8(m-1)$ . Handle $m=1$ before using a quadratic formula.
Answer

For $m>1$ : always two intersections. For $m<1$ , put $h=2-m$ and $d=\sqrt{8(1-m)}$ : zero if $|b-h|<d$ , one if $|b-h|=d$ , two if $|b-h|>d$ . For $m=1$ : zero if $b=1$ , and one otherwise.

Full solution

Equating the two formulas and clearing the nonzero denominator gives
$(m-1)x^2+(b-m)x-(b+1)=0.$
At $x=1$ the left side equals $-2$ , so the forbidden input can never be a root.
Counting roots of this polynomial therefore counts the curve-line intersections exactly.

If $m=1$ , the equation becomes
$(b-1)x-(b+1)=0.$
It has no solution when $b=1$ , and exactly one, $x=(b+1)/(b-1)$ , otherwise.
The excluded line $y=x+1$ is the slant asymptote: division gives $(x^2+1)/(x-1)=x+1+2/(x-1)$ , whose remainder term never vanishes.

If $m\ne1$ , the discriminant is
$(b-m)^2+4(m-1)(b+1)=(b+m-2)^2+8(m-1).$
For $m>1$ it is strictly positive, giving two intersections for every intercept.
For $m<1$ , set $h=2-m$ and
$d=\sqrt{8(1-m)}.$
The discriminant is $(b-h)^2-d^2$ , which is negative, zero, or positive according as $|b-h|<d$ , $=d$ , or $>d$ .
This proves the full classification.
In particular, slopes below $1$ have exactly two intercepts producing a single intersection, while slope $1$ has infinitely many such intercepts because the equation drops degree.

Answer

For $m>1$ : always two intersections. For $m<1$ , put $h=2-m$ and $d=\sqrt{8(1-m)}$ : zero if $|b-h|<d$ , one if $|b-h|=d$ , two if $|b-h|>d$ . For $m=1$ : zero if $b=1$ , and one otherwise.

Key idea

A discriminant classifies intersections only after degree changes and forbidden inputs have been checked.
Problem 7: Mark complete
Problem 8 Rational sums at unknown cubic roots

Difficulty: 3 of 3 stars, Deep challenge

Let $a,b,c$ be the three real zeros of $x^3-4x+1$ . Without finding them individually, evaluate
$S=\frac1{a^2+1}+\frac1{b^2+1}+\frac1{c^2+1},\qquad T=\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}.$
Explain why both sums are defined, and show the symmetric-sum calculations behind your answer.
1. Hint 1
  Vieta gives $a+b+c=0$ , $ab+bc+ca=-4$ , and $abc=-1$ . Compute $a^2+b^2+c^2$ and $a^2b^2+b^2c^2+c^2a^2$ .
2. Hint 2
  Use the common denominator $(a^2+1)(b^2+1)(c^2+1)$ . For the numerator of $T$ , separate the degree-five terms, the six degree-three terms, and the linear terms.
Answer

$S=35/26$ and $T=7/26$ .

Full solution

The zeros are real, so each denominator $a^2+1,b^2+1,c^2+1$ is positive.
Vieta gives $a+b+c=0$ , $ab+bc+ca=-4$ , and $abc=-1$ .
Consequently, $a^2+b^2+c^2=8$ and
$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)=16.$

The common denominator is $D=(a^2+1)(b^2+1)(c^2+1)=(abc)^2+(a^2b^2+b^2c^2+c^2a^2)+(a^2+b^2+c^2)+1=1+16+8+1=26$ .
The numerator of $S$ is $(a^2+1)(b^2+1)+(b^2+1)(c^2+1)+(c^2+1)(a^2+1)=16+2\cdot8+3=35$ .

For $T$ , the numerator is $a(b^2+1)(c^2+1)+b(c^2+1)(a^2+1)+c(a^2+1)(b^2+1)$ .
Its degree-five terms sum to
$abc(ab+bc+ca)=4.$
The six degree-three terms sum to
$(a+b+c)(ab+bc+ca)-3abc=3.$
Its linear terms sum to $0$ .
Thus the numerator is $7$ , proving both values.
This computation uses only symmetric information and never requires choosing an order for the three roots.

Answer

$S=35/26$ and $T=7/26$ .

Key idea

For rational sums at polynomial roots, expand the common denominator in symmetric pieces before attempting individual root formulas.
Problem 8: Mark complete
Problem 9 Where should the improvement go?

Difficulty: 3 of 3 stars, Deep challenge

An idealized process has three sequential stages. If nonnegative amounts $x,y,z$ of improvement are assigned to them, its total time is modeled by
$T(x,y,z)=\frac1{1+x}+\frac4{1+y}+\frac9{1+z}.$
Find the minimum possible time and every minimizing allocation in each case: (a) $x+y+z=2$ ; (b) $x+y+z=9$ . Prove global optimality without calculus. Explain why the optimal allocation gives nothing to one stage in one case but improves all three stages in the other.
1. Hint 1
  Set $u=1+x,v=1+y,w=1+z$ , so $u,v,w\ge1$ . The weighted reciprocal inequality follows from a square.
2. Hint 2
  For part (a), first use $4/v+9/w\ge25/(v+w)$ . Then compare $1/u+25/(5-u)$ to $29/4$ on $1\le u\le3$ . For part (b), use $(1/u+4/v+9/w)(u+v+w)\ge(1+2+3)^2$ .
Answer

(a) Minimum $29/4$ at $(x,y,z)=(0,3/5,7/5)$ only. (b) Minimum $3$ at $(x,y,z)=(1,3,5)$ only.

Full solution

For positive $p,q$ and nonnegative $A,B$ , direct expansion gives
$(A^2/p+B^2/q)(p+q)-(A+B)^2=(Aq-Bp)^2/(pq)\ge0.$
Equality requires $p:q=A:B$ when $A,B>0$ .
The three-term version follows by adding the analogous three pairwise squares; equality requires the three denominators to be proportional to the three numerator square roots.

For part (a), $u+v+w=5$ and each is at least $1$ , so
$1\le u\le3.$
The two-term inequality gives
$T\ge1/u+25/(5-u).$
Subtracting $29/4$ yields $(u-1)(29u-20)/(4u(5-u))\ge0$ on this interval.
Equality forces $u=1$ , followed by $v:w=2:3$ .
Since $v+w=4$ , this gives $v=8/5,w=12/5$ , and hence
$(x,y,z)=(0,3/5,7/5).$
All entries are feasible.

For part (b), $u+v+w=12$ , so the three-term inequality gives
$T\ge36/12=3.$
Equality requires $u:v:w=1:2:3$ , hence $(u,v,w)=(2,4,6)$ and
$(x,y,z)=(1,3,5).$
The lower bound $u\ge1$ prevents using the same unconstrained proportion in part (a): with total $5$ , it would require $u=5/6$ .
The boundary constraint is therefore essential, rather than a detail to check after optimizing.

Answer

(a) Minimum $29/4$ at $(x,y,z)=(0,3/5,7/5)$ only. (b) Minimum $3$ at $(x,y,z)=(1,3,5)$ only.

Key idea

A sharp unconstrained inequality may suggest an infeasible equality case; optimize again on the active boundary.
Problem 9: Mark complete
Problem 10 Zeros between vertical asymptotes

Difficulty: 3 of 3 stars, Deep challenge

Let $a_1<a_2<\cdots<a_n$ be real numbers, with $n\ge2$ , and let $w_1,\ldots,w_n$ be positive real weights. For a real constant $c$ , consider
$\frac{w_1}{x-a_1}+\frac{w_2}{x-a_2}+\cdots+\frac{w_n}{x-a_n}=c,\qquad x\ne a_1,\ldots,a_n.$
Prove that there is exactly one solution in every interval $(a_j,a_{j+1})$ . Determine how many additional real solutions occur outside $[a_1,a_n]$ for $c=0$ , $c>0$ , and $c<0$ , and locate them.

You may use that a continuous graph which takes values on both sides of a target height must reach that height somewhere between those inputs.
1. Hint 1
  On an interval containing no pole, compare the expression at two inputs by subtracting corresponding fractions.
2. Hint 2
  Near the left endpoint of an interior interval one term grows positively without bound, while near the right endpoint one grows negatively without bound. On each exterior interval, all denominators have the same sign.
Answer

Each of the $n-1$ interior intervals contains one solution. If $c=0$ , there are no others. If $c>0$ , there is one additional solution in $(a_n,\infty)$ ; if $c<0$ , one in $(-\infty,a_1)$ .

Full solution

Write the left side as $F(x)$ .
If $x<y$ belong to the same pole-free interval, then $w_j/(y-a_j)-w_j/(x-a_j)=-w_j(y-x)/((y-a_j)(x-a_j))<0$ : the two denominator factors have the same sign.
Adding proves $F(y)<F(x)$ , so each such interval contains at most one solution.

On $(a_j,a_{j+1})$ , the term $w_j/(x-a_j)$ becomes arbitrarily large and positive near the left endpoint, while the other terms stay bounded there.
Near the right endpoint, $w_{j+1}/(x-a_{j+1})$ becomes arbitrarily negative while the others stay bounded.
Thus some two interior inputs have values above and below any given $c$ .
Continuity gives a solution, and strict decrease gives uniqueness.

For $x>a_n$ , every term is positive.
The sum becomes arbitrarily large near $a_n$ and approaches $0$ as $x$ increases without bound.
Therefore it reaches each positive $c$ exactly once and never reaches a nonpositive $c$ .
For $x<a_1$ , every term is negative; the sum approaches $0$ from below as $x$ decreases without bound, and becomes arbitrarily negative near $a_1$ .
It reaches each negative $c$ exactly once and no nonnegative $c$ .
These intervals exhaust the permitted real domain, proving the full count.

Answer

Each of the $n-1$ interior intervals contains one solution. If $c=0$ , there are no others. If $c>0$ , there is one additional solution in $(a_n,\infty)$ ; if $c<0$ , one in $(-\infty,a_1)$ .

Key idea

Positive weights and pole locations can determine every real solution count without expanding a high-degree numerator.
Problem 10: Mark complete