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Graphing 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 Move the graph, then read its zeros

Difficulty: 1 of 3 stars, Stretch

The graph of $f$ consists exactly of the straight segments joining, in order, $(-4,1)$ , $(-1,4)$ , $(1,-2)$ , and $(3,2)$ . Both endpoints are included. Define $g(x)=1-2f(3-x)$ wherever the expression is defined.

Give the vertices of the graph of $g$ in left-to-right order, its domain and range, and all its zeros. Explain why simply transforming the zeros of $f$ would not answer the last question.

Text description of this figure

A coordinate grid with x running from -4 to 3 and y from -2 to 4, and the origin labeled 0. The graph of f is a broken line made of three straight segments. It starts at the point (-4, 1), rises to (-1, 4), falls to (1, -2), and rises again to end at (3, 2). Each of these four points is marked with a dot and labeled with its coordinates.

Builds on Graphs of Functions, Shifting Graphs, Stretching and Reflecting Graphs
1. Hint 1
  Track an old point $(u,v)$ by solving $3-x=u$ and calculating the new height $1-2v$ .
2. Hint 2
  The equation $g(x)=0$ asks where the original graph has height $1/2$ , not height $0$ .
Answer

Vertices $(0,-3),(2,5),(4,-7),(7,-1)$ ; domain $[0,7]$ , range $[-7,5]$ ; zeros $3/4$ and $17/6$ .

Full solution

An original point $(u,v)$ maps to $(3-u,1-2v)$ .
The four vertices therefore become $(7,-1),(4,-7),(2,5),(0,-3)$ .
Reversing their order gives the left-to-right list in the answer.
Straight segments remain straight under these coordinate changes.
The horizontal extent is $[0,7]$ , and the lowest and highest vertex heights are $-7$ and $5$ ; the connected segments attain every height between them.

For a zero, solve
$f(3-x)=1/2.$
The first original segment stays between heights $1$ and $4$ , so it contributes none.
On the middle segment, the equation of the line is $v=1-3u$ .
Its height is $1/2$ at $u=1/6$ , giving
$x=3-1/6=17/6.$
On the final segment, $v=2u-4$ , which equals $1/2$ at $u=9/4$ , giving $x=3/4$ .
Both old inputs lie in their segments, so both zeros are valid and the list is complete.

An old zero has height $v=0$ , so its new height is $1$ , not $0$ .
The vertical shift changes which horizontal level must be inspected before the input transformation is applied.

Answer

Vertices $(0,-3),(2,5),(4,-7),(7,-1)$ ; domain $[0,7]$ , range $[-7,5]$ ; zeros $3/4$ and $17/6$ .

Key idea

To find transformed zeros, first determine the required old output level, then transform its input locations.
Problem 1: Mark complete
Problem 2 When does order stop mattering?

Difficulty: 1 of 3 stars, Stretch

A function $f:\mathbb R\to\mathbb R$ has exactly one point of greatest height, at $(2,5)$ . Let $T_h$ shift its graph horizontally by $h$ units, where $h$ is any real number. Let $H_s$ stretch every horizontal coordinate by a factor $s>0$ , leaving heights unchanged.

Find all pairs $(h,s)$ for which applying $T_h$ and then $H_s$ gives exactly the same graph as applying $H_s$ and then $T_h$ . Prove both necessity and sufficiency.

If the unique-highest-point assumption is removed, must your condition still be necessary? Give a counterexample or a proof.

Builds on Shifting Graphs, Stretching and Reflecting Graphs, Graphs of Functions
1. Hint 1
  Follow the unique highest point through the two orders.
2. Hint 2
  The first order sends a horizontal coordinate $u$ to $s(u+h)$ ; the second sends it to $su+h$ .
Answer

Exactly $h=0$ or $s=1$ . Without the assumption the condition need not be necessary; a constant function is a counterexample.

Full solution

Both transformations preserve heights and are reversible, so the unique highest point stays unique.
In the order $T_h$ then $H_s$ , its horizontal coordinate becomes
$s(2+h)=2s+sh.$
In the other order it becomes $2s+h$ .
If the resulting graphs are identical, their unique highest points must coincide.
Hence $sh=h$ , or $(s-1)h=0$ , proving the necessary alternatives $h=0$ or $s=1$ .

To check sufficiency for the entire graph, track an arbitrary old point $(u,v)$ .
The two orders send it respectively to $(su+sh,v)$ and $(su+h,v)$ .
When $(s-1)h=0$ , these agree for every point, so the resulting graphs are identical.
Matching only one landmark gave necessity; the arbitrary-point calculation gives sufficiency.

Without a unique highest point, take $f(x)=5$ for every real $x$ .
Any horizontal shift or stretch leaves this horizontal line unchanged.
Thus, for example, $h=3$ and $s=2$ work even though neither of the stated alternatives holds.
A distinctive landmark rules out such extra symmetries.

Answer

Exactly $h=0$ or $s=1$ . Without the assumption the condition need not be necessary; a constant function is a counterexample.

Key idea

Track a unique geometric feature to prove necessity, and then track an arbitrary point to prove equality of whole graphs.
Problem 2: Mark complete
Problem 3 An inverse can meet off the diagonal

Difficulty: 1 of 3 stars, Stretch

The graph of $f:[0,4]\to[0,4]$ consists exactly of the segments from $(0,4)$ to $(2,3)$ and from $(2,3)$ to $(4,0)$ .

Prove that $f$ has an inverse. Find every point shared by the graphs of $f$ and $f^{-1}$ . Does every shared point lie on $y=x$ ? Justify completeness using the graph segments.

Text description of this figure

A coordinate grid in the first quadrant, with x and y each running from 0 to 4, and the origin labeled 0. The graph of f is two straight segments: a gentle fall from the point (0, 4) to (2, 3), then a steeper fall from (2, 3) to (4, 0). The three points are marked with dots and labeled with their coordinates.
1. Hint 1
  Reflect each vertex across $y=x$ to obtain the inverse graph.
2. Hint 2
  Use the breakpoints $x=2$ and $x=3$ to compare the two straight-line formulas on three intervals.
Answer

The shared points are $(0,4)$ , $(12/5,12/5)$ , and $(4,0)$ . Two of them are off the diagonal.

Full solution

The original graph is strictly decreasing and takes every output between $4$ and $0$ exactly once.
Thus it has an inverse on $[0,4]$ .
Reflecting its vertices gives the inverse graph with left-to-right vertices $(0,4),(3,2),(4,0)$ .

For $0\le x\le2$ , the original line has height $4-x/2$ , while the inverse line has height $4-2x/3$ .
Equality forces $x=0$ , giving $(0,4)$ .

For $2\le x\le3$ , the original height is $6-3x/2$ and the inverse height is still $4-2x/3$ .
Equality gives $2=5x/6$ , so $x=12/5$ .
The common height is also $12/5$ , and this input lies in the interval.

For $3\le x\le4$ , the inverse height is $8-2x$ .
Equating this to $6-3x/2$ gives $x=4$ , hence $(4,0)$ .
The three intervals cover the common domain, so the list is complete.

The off-diagonal points occur because $f(0)=4$ and $f(4)=0$ : applying $f$ twice returns to the starting input even though one application changes it.
Reflection across $y=x$ does not force two reflected graphs to intersect only on the mirror line.

Solid: $f$ . Dashed: $f^{-1}$

Answer

The shared points are $(0,4)$ , $(12/5,12/5)$ , and $(4,0)$ . Two of them are off the diagonal.

Key idea

An inverse-graph intersection can represent a two-step return, so check the monotonic direction before imposing $x=y$ .
Problem 3: Mark complete
Problem 4 Area exchanged by reflection

Difficulty: 2 of 3 stars, Challenge

The graph of $f:[0,6]\to[0,6]$ consists exactly of the segments joining $(0,0)$ , $(1,3)$ , $(3,4)$ , and $(6,6)$ .

Find the exact area enclosed by the graphs of $f$ and $f^{-1}$ . Justify which graph is higher between their endpoints and explain how reflection across $y=x$ lets you calculate the area without first finding formulas for the inverse.

Text description of this figure

A coordinate grid in the first quadrant, with x and y each running from 0 to 6. The graph of f is a broken line of three straight segments that rises from the point (0, 0) to (1, 3), then to (3, 4), then to (6, 6). The four points are marked with dots and labeled with their coordinates.
1. Hint 1
  Compare each original segment with the diagonal of the $6$ by $6$ square.
2. Hint 2
  Reflection sends the region below the graph of $f$ to the region to the left of the inverse graph. Relate the areas under the two graphs to the square.
Answer

The enclosed area is $11$ square units.

Full solution

The graph is strictly increasing and runs from $(0,0)$ to $(6,6)$ , so it has an inverse on the same interval.
At the two internal vertices, $y-x$ is $2$ and $1$ , and it is zero at the endpoints.
Since $y-x$ changes linearly on each segment, the graph is strictly above $y=x$ for every interior input.
Its reflected inverse is strictly below the diagonal there.
Consequently the two graphs enclose a single region.

Let $A$ be the area under $f$ in the square $[0,6]\times[0,6]$ , and let $B$ be the area under its inverse.
Reflection in $y=x$ takes the region below $f$ to the region above the inverse, so $A=36-B$ .
This uses the fact that reflection preserves area; it does not require inverse formulas.

Using one triangle and two trapezoids, the original area is $A=\tfrac12(1)(3)+\tfrac12(3+4)(2)+\tfrac12(4+6)(3)=3/2+7+15=47/2$ .
Thus
$B=36-47/2=25/2.$

Because $f$ is the higher graph throughout the interior, the requested area is
$A-B=47/2-25/2=11.$
Subtracting areas works here because the graphs do not cross between the endpoints.

Shaded area: $11$ square units

Answer

The enclosed area is $11$ square units.

Key idea

Use geometric symmetry to exchange a difficult area with a complementary area that is easy to calculate.
Problem 4: Mark complete
Problem 5 One final graph, many transformations

Difficulty: 2 of 3 stars, Challenge

Let $f(x)=x^2-4x+1$ on all real numbers. Find every quadruple of real numbers $(a,b,c,d)$ with $b\ne0$ such that the transformed graph $y=a f(bx+c)+d$ is exactly the graph $y=(x-3)^2+2$ .

Explain why the final graph does not uniquely determine its horizontal and vertical stretch factors.

Builds on Shifting Graphs, Stretching and Reflecting Graphs
1. Hint 1
  Put the original parabola in vertex form before composing.
2. Hint 2
  The final leading coefficient and vertex impose three conditions, while the horizontal factor $b$ can remain free.
Answer

For any real $b\ne0$ , take $a=1/b^2$ , $c=2-3b$ , and $d=2+3/b^2$ . These are all quadruples.

Full solution

The original function is
$f(u)=(u-2)^2-3.$
Thus the transformed equation is
$y=a(bx+c-2)^2+d-3a.$
Its coefficient of $x^2$ is $ab^2$ .
The target coefficient is $1$ , so $a=1/b^2>0$ .
In particular, a zero or negative vertical coefficient cannot work.

The transformed vertex occurs where $bx+c-2=0$ .
Since the target vertex has input $3$ , we need $c-2=-3b$ , or $c=2-3b$ .
Its height is $d-3a$ , which must equal $2$ , giving $d=2+3/b^2$ .
These conditions were forced by equality of the graphs.

Conversely, for any nonzero real $b$ , the displayed choices give $a(bx+c-2)^2=(1/b^2)[b(x-3)]^2=(x-3)^2$ and $d-3a=2$ .
Hence every quadruple in the family works.

The free parameter represents a trade between horizontal and vertical scaling: squaring multiplies a horizontal input factor $b$ into a height factor $b^2$ , which the coefficient $a=1/b^2$ exactly cancels.
The sign of $b$ can also reverse horizontal orientation without changing the squared expression.
Thus one final parabola can arise through infinitely many transformation descriptions.

Answer

For any real $b\ne0$ , take $a=1/b^2$ , $c=2-3b$ , and $d=2+3/b^2$ . These are all quadruples.

Key idea

A transformation description need not be unique when the original graph has scaling or reflection symmetries.
Problem 5: Mark complete
Problem 6 A moving copy of a tent

Difficulty: 2 of 3 stars, Challenge

The graph of $f$ consists exactly of the segments from $(0,0)$ to $(2,6)$ and from $(2,6)$ to $(5,0)$ , with domain $[0,5]$ . For each real $h\ge0$ , shift this graph $h$ units to the right, obtaining $y=f(x-h)$ on $[h,h+5]$ .

For every $h\ge0$ , determine how many points the shifted graph shares with the original. When the number is finite and positive, give the intersection coordinates in terms of $h$ . Prove completeness.

Text description of this figure

A pair of labeled axes with no grid. The graph, labeled y equals f of x, is two straight segments: it rises from the point (0, 0) to a peak at (2, 6), then falls to the point (5, 0) on the x axis. The three points are marked with dots and labeled with their coordinates.

Builds on Shifting Graphs
1. Hint 1
  If $h>0$ , the shifted graph uses the smaller old input $x-h$ . Equal heights cannot come from two inputs on the same strictly monotone side of the tent.
2. Hint 2
  A meeting must use the falling side of the original tent and the rising side of the shifted tent. Their heights are $10-2x$ and $3(x-h)$ .
Answer

For $h=0$ , the graphs coincide. For $0<h\le5$ , there is exactly one shared point: $(2+3h/5,\,6-6h/5)$ . For $h>5$ , there are none.

Full solution

At $h=0$ , the graphs are identical and share infinitely many points.
If $h>5$ , their domains $[0,5]$ and $[h,h+5]$ are disjoint, so no intersection is possible.

Now let $0<h\le5$ .
At a common input, $x-h<x$ .
If both old inputs lie on the rising side, strict increase makes their outputs unequal.
If both lie on the falling side, strict decrease does the same.
The only remaining possibility places $x-h$ on the rising side and $x$ on the falling side.
It requires $0\le x-h\le2$ and
$2\le x\le5.$

The rising-side equation is $f(u)=3u$ ; the falling-side equation is
$f(u)=10-2u.$
Thus a common point must satisfy $10-2x=3(x-h)$ , giving $x=2+3h/5$ and $y=6-6h/5$ .

For every $0<h\le5$ , this candidate satisfies $2\le x\le5$ and
$x-h=2-2h/5\in[0,2].$
It therefore belongs to the correct segments and works.
The linear equation has only one solution, so it is the unique meeting point.
At $h=5$ the formula gives $(5,0)$ , the single shared endpoint of the domains.

Answer

For $h=0$ , the graphs coincide. For $0<h\le5$ , there is exactly one shared point: $(2+3h/5,\,6-6h/5)$ . For $h>5$ , there are none.

Key idea

Use monotonicity to determine which graph pieces can meet before solving their equations.
Problem 6: Mark complete
Problem 7 Tilting a graph through three zeros

Difficulty: 2 of 3 stars, Challenge

The graph of $f$ consists exactly of the segments joining $(-3,0)$ , $(-1,3)$ , $(1,-1)$ , and $(4,2)$ , with endpoints included. For a real parameter $k$ , define $g_k(x)=f(x)+kx$ on $[-3,4]$ .

Find all $k$ for which $g_k$ has exactly three distinct zeros. Account carefully for zeros at vertices and at domain endpoints; a shared vertex counts only once.

Text description of this figure

A coordinate grid with x running from -3 to 4 and y from -1 to 3, and the origin labeled 0. The graph of f is a broken line of three straight segments. It starts on the x axis at the point (-3, 0), rises to (-1, 3), falls to (1, -1), and rises again to end at (4, 2). Each of these four points is marked with a dot and labeled with its coordinates.

Builds on Graphs of Functions
1. Hint 1
  Adding $kx$ keeps each segment straight and changes only its vertex heights.
2. Hint 2
  The new heights, in order, are $-3k$ , $3-k$ , $k-1$ , and $4k+2$ . For three distinct zeros, each of the three segments must supply one.
Answer

Exactly $0\le k<1$ .

Full solution

The new vertex heights are $A=-3k$ , $B=3-k$ , $C=k-1$ , and $D=4k+2$ .
No adjacent pair can both be zero: the required values of $k$ would conflict in each case.
Thus no whole segment lies on the axis, and each segment contains at most one zero.

Three distinct zeros therefore require one on each segment.
Neither internal vertex can be zero, since a zero shared by two segments would leave at most two distinct zeros in total.
In particular, the middle segment must cross strictly through the axis: $BC<0$ .
This means $k<1$ or $k>3$ .

If $k>3$ , both $A$ and $B$ are negative, so the first segment has no zero.
This case fails.
For $k<1$ , we have $B>0$ and $C<0$ .
The first segment has a zero precisely when $A\le0$ , or $k\ge0$ ; equality allows the left domain endpoint.
The last segment has a zero precisely when $D\ge0$ , or $k\ge-1/2$ .
Combining these gives $0\le k<1$ .

Conversely, throughout this interval the first segment meets the axis, the middle crosses it, and the last crosses it.
None of these zeros is a shared internal vertex, so there are exactly three.
At $k=1$ the middle and last zeros merge at the vertex $(1,0)$ , explaining why the upper endpoint is excluded.

Answer

Exactly $0\le k<1$ .

Key idea

Count distinct intersections by signs at landmarks, treating shared vertices separately from crossings.
Problem 7: Mark complete
Problem 8 Two symmetries force a repeating graph

Difficulty: 3 of 3 stars, Deep challenge

The graph of a function $f:\mathbb R\to\mathbb R$ is unchanged by reflection across the vertical line $x=3$ and by a half-turn about $(1,2)$ . A half-turn sends a point $(u,v)$ to $(2-u,4-v)$ . Also $f(0)=1$ .

(a) Prove that shifting the graph $8$ units horizontally leaves it unchanged.

(b) Find $f(2026)$ .

(c) Could shifting it $4$ units horizontally also leave it unchanged? Justify your answer. Do not assume the graph is a parabola or a straight line.

Builds on Shifting Graphs, Stretching and Reflecting Graphs
1. Hint 1
  Translate the two geometric symmetries into identities relating function values.
2. Hint 2
  Compose the two point transformations. One composition shifts the input by $4$ while reflecting the output about height $2$ .
Answer

(a) $f(x+8)=f(x)$ for every real $x$ . (b) $f(2026)=3$ . (c) No: $f(4)=3\ne f(0)$ .

Full solution

Reflection across $x=3$ gives
$f(6-x)=f(x).$
The half-turn gives
$f(2-x)=4-f(x).$
Apply the reflection identity to the input $2-x$ to obtain
$f(x+4)=f(2-x)=4-f(x).$
Applying this new relation twice yields $f(x+8)=4-[4-f(x)]=f(x)$ for every real $x$ .
Thus an $8$ -unit shift preserves the graph.

The half-turn relation at $x=0$ gives
$f(2)=4-f(0)=3.$
Since $2026=2+253\cdot8$ , repeated use of the period gives
$f(2026)=f(2)=3.$

The four-unit relation gives
$f(4)=4-f(0)=3.$
This differs from $f(0)=1$ , so a four-unit shift cannot preserve the graph.
Proving that $8$ is a period does not by itself prove it is the smallest positive period; that stronger assertion was not required.

The assumptions are consistent.
One example is obtained by joining $(-1,0)$ to $(3,4)$ to $(7,0)$ with straight segments and repeating that pattern every $8$ units.
Its reflected and half-turned copies agree with the same repeating graph, and its value at $0$ is $1$ .

Answer

(a) $f(x+8)=f(x)$ for every real $x$ . (b) $f(2026)=3$ . (c) No: $f(4)=3\ne f(0)$ .

Key idea

Compose graph symmetries algebraically; two reflections or rotations can force a translation relation.
Problem 8: Mark complete
Problem 9 Intersections with a moving inverse

Difficulty: 3 of 3 stars, Deep challenge

For each real parameter $c$ , define $f_c(x)=(x-c)^2$ on the domain $[c,\infty)$ .

Determine, for every real $c$ , all points shared by the graphs of $f_c$ and its inverse, and give the number of such points. Your argument must justify why no off-diagonal intersections are possible and must respect both graph domains.

Builds on Shifting Graphs
1. Hint 1
  The restricted function is strictly increasing. If $f_c(u)=v$ and $f_c(v)=u$ , compare the orders of $u$ and $v$ .
2. Hint 2
  After proving a common point lies on $y=x$ , put $t=x-c\ge0$ . The resulting equation is quadratic in $t$ .
Answer

No points for $c<-1/4$ ; one point $(1/4,1/4)$ for $c=-1/4$ ; two for $-1/4<c\le0$ ; one for $c>0$ . The possible diagonal coordinates are $s_\pm=(1+2c\pm\sqrt{1+4c})/2$ , retaining both signs in the two-point case and only $+$ when $c>0$ .

Full solution

On $[c,\infty)$ , the quantity $x-c$ is nonnegative and strictly increases, so $f_c$ is strictly increasing with range $[0,\infty)$ .
Its inverse is $c+\sqrt{x}$ on $[0,\infty)$ .
If $(u,v)$ belongs to both graphs, then $f_c(u)=v$ and $f_c(v)=u$ , with both inputs allowed.
If $u<v$ , increasing behavior would imply $v=f_c(u)<f_c(v)=u$ , a contradiction.
The case $u>v$ fails similarly.
Thus every shared point is diagonal.

Put $u=v=x$ and $t=x-c\ge0$ .
Then $x=(x-c)^2$ is equivalent to $t+c=t^2$ , or
$(t-1/2)^2=c+1/4.$
Any valid solution automatically has $x=t^2\ge0$ , so it lies in the inverse domain as well as the original domain.

For $c<-1/4$ there are no real $t$ .
At $c=-1/4$ , $t=1/2$ gives $x=1/4$ .
For larger $c$ , the candidates are
$t_\pm=(1\pm\sqrt{1+4c})/2.$
Both are nonnegative when $-1/4<c\le0$ , giving two distinct points.
When $c>0$ , only $t_+$ is nonnegative, giving one.

Finally $x=t+c$ yields the stated coordinates $s_\pm$ .
Each retained value solves the original fixed-point equation on both domains, so every listed diagonal point really is shared.

Answer

No points for $c<-1/4$ ; one point $(1/4,1/4)$ for $c=-1/4$ ; two for $-1/4<c\le0$ ; one for $c>0$ . The possible diagonal coordinates are $s_\pm=(1+2c\pm\sqrt{1+4c})/2$ , retaining both signs in the two-point case and only $+$ when $c>0$ .

Key idea

An increasing function has no nontrivial two-step cycles; use that order fact before solving inverse-graph equations.
Problem 9: Mark complete
Problem 10 Two graph rules determine every point

Difficulty: 3 of 3 stars, Deep challenge

Find all strictly increasing functions $f:\mathbb R\to\mathbb R$ whose graphs have both of these properties: shifting every point right by $1$ and up by $2$ leaves the graph unchanged; multiplying both coordinates of every point by $2$ also leaves the graph unchanged.

Prove that your list is complete without assuming a formula, continuity, or differentiability for $f$ . You may use the elementary fact that powers of $2$ eventually exceed any fixed positive real number.

Builds on Shifting Graphs, Stretching and Reflecting Graphs
1. Hint 1
  Turn the geometric rules into identities for $f(x+1)$ and $f(2x)$ . Determine $f$ at integer inputs first.
2. Hint 2
  For a positive integer $m$ , choose consecutive integers $n,n+1$ bracketing $2^m x$ . Increasing behavior traps $f(x)$ between two values whose gap shrinks as $m$ grows.
Answer

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

Full solution

The graph rules give $f(x+1)=f(x)+2$ and
$f(2x)=2f(x).$
The second identity at $0$ forces $f(0)=0$ .
The first then gives $f(n)=2n$ for every integer $n$ , using repeated shifts in either direction.
Repeated doubling gives $f(2^m x)=2^m f(x)$ for every positive integer $m$ .

Fix an arbitrary real input $x$ .
For any positive integer $m$ , choose an integer $n$ such that
$n\le2^m x<n+1.$
Since $f$ is strictly increasing,

$2n\le f(2^m x)<2n+2,\qquad\text{so}\qquad\frac{2n}{2^m}\le f(x)<\frac{2n+2}{2^m}.$

The same two bounds also contain $2x$ , by multiplying the bracketing inequality by $2/2^m$ .
Thus the distance between $f(x)$ and $2x$ is less than $2/2^m$ , for every positive integer $m$ .

If that distance were a positive number $d$ , choose $m$ so large that $2^m>2/d$ .
The distance would then be less than $2/2^m<d$ , a contradiction.
Therefore $f(x)=2x$ .
Since $x$ was arbitrary, the rule holds everywhere.
Conversely, $f(x)=2x$ is strictly increasing and its graph is unchanged by both specified transformations, proving existence and uniqueness.

Answer

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

Key idea

Order can trap an unknown graph between arbitrarily close known points, forcing exact values without a continuity assumption.
Problem 10: Mark complete