Inverse functions and original restrictions
Reverse an original mapping only where the recovery is unique, retain its actual image and restrictions, and distinguish an inverse from a reciprocal.
Section 4 · Functions (all14rows linked;self-check practice)
Exam scope and assessment notes
Original selected inverse teaching,with four stages,eight tasks,nine reversed branches and27 named inverse-input checks. Retain the original domain,actual image,stated target,one-to-one evidence,original holes,branch signs,return checks and swapped units. A larger target is not automatically the original image. An inverse undoes the mapping;it is not the reciprocal. Endpoint sets remain exact rational intervals. Selected simple square-root inverse values are compared exactly against the original branch and returned through the original equation;general symbolic forms and arbitrary inverse solving remain outside this bounded self-check. A squared return alone cannot choose an inverse side or change the original domain. A complete incorrect attempt may study the explained reference without passing fixed comparisons. All14Functions source rows have reviewed structural teaching/practice links;other Mathematics B source gaps remain explicit in the current course map;general symbolic inverse alternatives need human review and the remaining Pearson curriculum is incomplete. Written derivations,reversed equations and whole-domain proofs need human review. No official marks,mastery,specialist approval or formal assessment readiness. Single Higher-tier linear4MB1,entry context2026/27 only. ISBN9780435044107 is metadata only;no verified book body,chapter or Hub mapping and no MathematicsA,Foundation or modular inheritance
Before you begin
- Functions,mappings,actual domains and ranges,composition,interval sets,principal square roots and rearranging equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does an inverse function do?
What you will learn
- Interpret an inverse as recovering an original input from its output, rather than taking a reciprocal.
- Determine whether the whole original mapping is one-to-one and give a collision when it is not.
- Derive selected inverse rules using the original domain to retain signs, excluded values and endpoint decisions.
- Distinguish the actual image from a larger stated target and reverse the practical input/output units.
Each stage starts with blank construction and answer fields. Changing stage begins a fresh attempt.
Inverse functions and original restrictions: Worked
Keep the whole original domain,image and stated target. Reverse each justified input/output piece,or give a collision showing why the whole reverse relation is not a function. Then enter a separate final answer.
Use intervals such as [0,2),unions with U,singleton {2},and -inf or inf for infinite bounds. Type none for an empty set,a dimensionless unit or no available inverse value. Type not-a-function in the inverse-set fields when there is no inverse function on the whole original image;this differs from an empty inverse domain.
The inverse rule menu shows the formula template for your parameters. The square-root branch templates use the original a,h,k and your chosen sign. Add your own justified branches;inverse means undoing the mapping,not taking its reciprocal.
Endpoint sets remain exact rational intervals. Selected simple square-root inverse values can be entered exactly using sqrt(...). A real irrational value is still defined;do not replace it with none or a decimal approximation. General symbolic forms need teacher review. Your derivations,reversed equations and whole-domain proofs always need human review;fixed comparisons do not approve those sentences.
Find the inverse on the actual image of each original rule. First use f(x)=3x−5 with domain and target R. Then use g(x)=(x²−9)/(x−3),with original domain R except3 and stated target R. Reverse the original input/output pairs,retain each missing value,and distinguish an inverse on the actual image from an inverse defined on every member of a larger stated target.
Task 1: reverse the original mapping
Original rule: y=(3)*x+(-5).
Original input restriction: (-inf,inf). Retain additional denominator and radical conditions from the original rule.
Stated original target: (-inf,inf). A target is not automatically the actual image.
Named inverse-input probes: -5, 1, 4. Original input unit: dimensionless. Original output unit: dimensionless.
Inspect my entered construction
This display contains only your current entries. It supplies no inverse formula,branch,collision witness or return value.
My original domain,image and target
- My original allowed real input set
- Not entered
- My actual original image
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- The stated original target set
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My original-function and inverse decisions
- Is the original function one-to-one on its whole allowed domain?
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- Does the actual original image fill the stated target?
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- Is there an inverse function on the whole actual image?
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- Is there an inverse function on every member of the stated target?
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My inverse input and output sets
- My inverse domain (or not-a-function)
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- My inverse range (or not-a-function)
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My original and reversed units
- My original input unit (or none)
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- My original output unit (or none)
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- My inverse input unit (or none)
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- My inverse output unit (or none)
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My inverse and original-rule return trace for input -5
- My inverse input classification
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- Is this input admitted by the inverse function?
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- My exact recovered original input (or none)
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- Does the recovered value belong to the original domain?
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- My original function output at that recovered input (or none)
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- Explain the original-rule return check or why this is not an exact rational inverse value
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My inverse and original-rule return trace for input 1
- My inverse input classification
- Not entered
- Is this input admitted by the inverse function?
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- My exact recovered original input (or none)
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- Does the recovered value belong to the original domain?
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- My original function output at that recovered input (or none)
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- Explain the original-rule return check or why this is not an exact rational inverse value
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My inverse and original-rule return trace for input 4
- My inverse input classification
- Not entered
- Is this input admitted by the inverse function?
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- My exact recovered original input (or none)
- Not entered
- Does the recovered value belong to the original domain?
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- My original function output at that recovered input (or none)
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- Explain the original-rule return check or why this is not an exact rational inverse value
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My reversed relation and whole-domain reasoning
- Derive the inverse without changing the original domain
- Not entered
- Explain uniqueness or collision,actual image,target,branches and endpoint restrictions
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Construction status: Not recorded
My separate final answer
Enter your inverse decisions,formula,sets,units and named inverse values independently. Your construction entries are not copied into these answer fields.
Task 2: reverse the original mapping
Original rule: y=((1)*x+(3))*(x-(3))/(x-(3)).
Original input restriction: (-inf,inf). Retain additional denominator and radical conditions from the original rule.
Stated original target: (-inf,inf). A target is not automatically the actual image.
Named inverse-input probes: 3, 6, 9. Original input unit: dimensionless. Original output unit: dimensionless.
Inspect my entered construction
This display contains only your current entries. It supplies no inverse formula,branch,collision witness or return value.
My original domain,image and target
- My original allowed real input set
- Not entered
- My actual original image
- Not entered
- The stated original target set
- Not entered
My original-function and inverse decisions
- Is the original function one-to-one on its whole allowed domain?
- Not entered
- Does the actual original image fill the stated target?
- Not entered
- Is there an inverse function on the whole actual image?
- Not entered
- Is there an inverse function on every member of the stated target?
- Not entered
My inverse input and output sets
- My inverse domain (or not-a-function)
- Not entered
- My inverse range (or not-a-function)
- Not entered
My original and reversed units
- My original input unit (or none)
- Not entered
- My original output unit (or none)
- Not entered
- My inverse input unit (or none)
- Not entered
- My inverse output unit (or none)
- Not entered
My inverse and original-rule return trace for input 3
- My inverse input classification
- Not entered
- Is this input admitted by the inverse function?
- Not entered
- My exact recovered original input (or none)
- Not entered
- Does the recovered value belong to the original domain?
- Not entered
- My original function output at that recovered input (or none)
- Not entered
- Explain the original-rule return check or why this is not an exact rational inverse value
- Not entered
My inverse and original-rule return trace for input 6
- My inverse input classification
- Not entered
- Is this input admitted by the inverse function?
- Not entered
- My exact recovered original input (or none)
- Not entered
- Does the recovered value belong to the original domain?
- Not entered
- My original function output at that recovered input (or none)
- Not entered
- Explain the original-rule return check or why this is not an exact rational inverse value
- Not entered
My inverse and original-rule return trace for input 9
- My inverse input classification
- Not entered
- Is this input admitted by the inverse function?
- Not entered
- My exact recovered original input (or none)
- Not entered
- Does the recovered value belong to the original domain?
- Not entered
- My original function output at that recovered input (or none)
- Not entered
- Explain the original-rule return check or why this is not an exact rational inverse value
- Not entered
My reversed relation and whole-domain reasoning
- Derive the inverse without changing the original domain
- Not entered
- Explain uniqueness or collision,actual image,target,branches and endpoint restrictions
- Not entered
Construction status: Not recorded
My separate final answer
Enter your inverse decisions,formula,sets,units and named inverse values independently. Your construction entries are not copied into these answer fields.
Record every current original inverse construction table.
Undo a mapping, not an output value
The inverse reverses original input/output pairs. If f sends a to b, an inverse function sends b back to a. The notation f⁻¹ does not mean 1/f: a reciprocal changes an output number, whereas an inverse recovers the input that produced it.
For an inverse to be a function on the whole actual image, each original output must identify exactly one original input. This is the one-to-one condition. Checking a few named inputs can illustrate the recovery, but it does not establish one-to-one behavior for the entire original domain.
Retain the original domain and actual image
The inverse input set is the actual image of the original function. Its output set is the original domain. Preserve every original denominator exclusion, radical condition and stated restriction before simplifying or reversing a rule. Algebra alone does not restore a missing original input or an output it could not produce.
A separately stated target may contain values the original function never reaches. A one-to-one original mapping can have an inverse on its actual image while having no inverse defined on every member of that larger target. State which input set the inverse uses instead of inventing a recovery for an unreached target value.
Choose signs from the whole original mapping
Solving an equation involving a square may produce two possible signs. The original domain decides which branch can recover its inputs. A quadratic on a whole symmetric interval is not one-to-one; changing it to one side is a declared restriction, not an automatic correction to the original task.
Some disconnected original domains use both sides while keeping their image pieces disjoint. Their inverse can require different signs on different image pieces. Retain every piece and its endpoints, and explain why no original output is shared by distinct allowed inputs.
Check recovery and practical meaning
For an allowed original input x, applying the original rule and then its inverse should recover x. For an allowed inverse input y in the actual image, applying the inverse and then the original should recover y. Both identities are restricted to their correct input sets; an expression that can be evaluated elsewhere does not extend the mapping.
In a practical model, the original output unit becomes the inverse input unit and the original input unit becomes the inverse output unit. Keep the original operating restrictions and attained endpoints. This proposed bounded self-check supports selected rational interval endpoints and exact rational values; a real irrational recovery stays defined and should be written exactly. This selected self-check compares simple square-root candidates to the original inverse branch and checks their original-rule return;general symbolic forms need teacher review instead of a rounded substitute.
Sources for this lesson
Pause and explain
An original quadratic maps both−1and1to1. What does this show about reversing that whole mapping?
Worked example
Find the inverse on the actual image of each original rule. First use f(x)=3x−5 with domain and target R. Then use g(x)=(x²−9)/(x−3),with original domain R except3 and stated target R. Reverse the original input/output pairs,retain each missing value,and distinguish an inverse on the actual image from an inverse defined on every member of a larger stated target.
Show the worked solution
- For f, write y=3x−5 and solve for the original input: x=(y+5)/3. Reversing the labels gives f⁻¹(x)=(x+5)/3. The original rule maps R one-to-one onto R, so both the inverse domain and inverse range are R.
- The named inverse inputs −5, 1 and 4 recover original inputs 0, 2 and 3. Substituting each recovered input back into the original rule returns the named inverse input. The inverse is not the reciprocal 1/f(x): for example, f⁻¹(1)=2, whereas 1/f(1)=−1/2.
- For g, the original denominator excludes x=3. Simplifying to x+3 is valid only on that original domain. Its actual image is R except 6, since output 6 would require the forbidden original input 3. Each allowed original output y is reversed by x=y−3.
- Thus the inverse on the actual image is g⁻¹(y)=y−3, with inverse domain R except 6 and inverse range R except 3. The named inverse inputs 3 and 9 give 0 and 6; input 6 is outside the inverse domain. The separately stated target R is larger than the actual image: there is no inverse defined on every member of that whole target without changing the task.
Answer f⁻¹(x)=(x+5)/3 on R. g⁻¹(x)=x−3 on R except 6, with output set R except 3. The original g is one-to-one but does not reach its entire stated target R.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Both rules have formula(x−1)²+2 and target[2,∞). One original domain is[1,∞);the other is(−∞,1]. Find each inverse and its actual domain and range. Use the original side restriction to choose the square-root sign;do not change either original domain silently. For the right-hand rule,also recover the input for original output4 exactly,without rounding or declaring it unavailable.
Give me a hint
Solve for the original input,then use its original side of x=1 to choose the sign.
Compare my reasoning
- Let y=(x−1)²+2. Solving the squared relation gives x−1=±√(y−2). This relation alone does not choose the inverse sign. Use the original domain before reversing the mapping.
- On the original domain x≥1, the quantity x−1 is nonnegative. Choose x=1+√(y−2). The original image is [2,∞), so the inverse domain is [2,∞) and the inverse range is [1,∞). Named inverse inputs 2, 3 and 6 give 1, 2 and 3. The additional inverse input4 gives1+√2 exactly;this is a real allowed original input,not an undefined value or a rounded decimal. Substituting it into the original quadratic gives(√2)²+2=4,while the original right-hand restriction selects the positive square-root branch.
- On the original domain x≤1, the quantity x−1 is nonpositive. Choose x=1−√(y−2). The original image is again [2,∞), but the inverse range is now (−∞,1]. Named inverse inputs 2, 3 and 6 give 1, 0 and −1.
- Each restricted original rule is one-to-one and reaches its stated target. Substituting the chosen inverse branch into the original quadratic returns the allowed inverse input. Substituting the original quadratic into its inverse recovers the original input only when that input belongs to the stated original side.
Right-side restriction: inverse 1+√(x−2), domain [2,∞), range [1,∞);inverse input4 gives1+√2 exactly. Left-side restriction: inverse 1−√(x−2), domain [2,∞), range (−∞,1].
Look for these in your work
- Retain the original formula, original admissible domain and separately stated target.
- Establish whether distinct original inputs can share an output; give an actual collision when they can.
- Use the actual original image as the inverse domain and the original domain as the inverse range.
- Reverse the actual branch or pair mapping and retain the correct sign, exclusions and endpoint decisions.
- Check the recovery on allowed inputs and swap the input/output units. Written derivations still need human review.
For f(x)=x² on[-2,2] with target[0,4],decide whether reversing the whole mapping produces a function and give a collision when it does not. Separately,g(x)=x² has original domain[-3,−2]∪[0,1] and target[0,1]∪[4,9]. Determine whether it is one-to-one,and,if so,construct its inverse on the two actual output pieces. A few inverse values do not replace the whole-domain argument.
Give me a hint
Look for repeated original outputs. For the disconnected domain,compare the two entire image pieces before choosing each inverse branch.
Compare my reasoning
- For f on [−2,2], distinct inputs −2 and 2 both give output 4; inputs −1 and 1 both give output 1. Reversing the entire mapping would therefore give more than one output for some inverse inputs. There is no inverse function for that whole original mapping.
- The reversed relation has the unique pair 0↦0 at one value, but this does not make the entire reversed relation a function. A restricted original domain could produce an inverse, but that would be a declared change to this task, not an automatic repair.
- For g, the original branch [−3,−2] maps one-to-one onto [4,9], and [0,1] maps one-to-one onto [0,1]. These two image sets are disjoint, so no output is shared by the two original branches. The whole restricted mapping is one-to-one despite using both sides of a quadratic.
- Its inverse is √y for 0≤y≤1 and −√y for 4≤y≤9. The inverse domain is [0,1]∪[4,9], and the inverse range is [−3,−2]∪[0,1]. Named inverse inputs 0, 1, 4 and 9 give 0, 1, −2 and −3. Input 2 is outside the actual image, even though √2 is a real number.
f has no inverse function on its whole stated domain. g has the two-piece inverse √x on [0,1] and −√x on [4,9], with inverse range [−3,−2]∪[0,1].
Look for these in your work
- Retain the original formula, original admissible domain and separately stated target.
- Establish whether distinct original inputs can share an output; give an actual collision when they can.
- Use the actual original image as the inverse domain and the original domain as the inverse range.
- Reverse the actual branch or pair mapping and retain the correct sign, exclusions and endpoint decisions.
- Check the recovery on allowed inputs and swap the input/output units. Written derivations still need human review.
A hypothetical reading is r(x)=2√(x−1)+3 reading units for dimensionless settings1≤x<10. Recover the setting from an allowed reading and retain the excluded upper boundary. Separately,a12km trip has speed s(t)=12/t km/h for2≤t<6 hours. Recover the permitted time from a speed,retaining both endpoint decisions and reversing the input/output units.
Give me a hint
Find the actual original output set first;it becomes the inverse input set. Reverse the units as well as the equation.
Compare my reasoning
- For the hypothetical calibration, the original settings [1,10) give readings [3,9). Solve r=2√(x−1)+3: (r−3)/2=√(x−1), then x=1+((r−3)/2)². Retain 3≤r<9 before squaring.
- The inverse takes reading units as input and returns a dimensionless setting. Readings 3 and 5 give settings 1 and 2. Reading 9 is excluded because it would require original setting 10. A value below 3 cannot become an inverse input merely because the squared expression can be evaluated there.
- For the journey, solve v=12/t to obtain t=12/v. Original times [2,6) hours give speeds (2,6] km/h. These speed values form the inverse domain; the recovered times [2,6) hours form its range.
- Inverse input 2km/h is excluded because it would require 6 hours. Inputs 3km/h and 6km/h recover 4 hours and 2 hours. The same reciprocal-looking algebra does not make the original and inverse domains or units identical.
Calibration inverse: 1+((r−3)/2)² for 3≤r<9, returning settings [1,10). Journey inverse: 12/v hours for 2<v≤6 km/h, returning times [2,6).
Look for these in your work
- Retain the original formula, original admissible domain and separately stated target.
- Establish whether distinct original inputs can share an output; give an actual collision when they can.
- Use the actual original image as the inverse domain and the original domain as the inverse range.
- Reverse the actual branch or pair mapping and retain the correct sign, exclusions and endpoint decisions.
- Check the recovery on allowed inputs and swap the input/output units. Written derivations still need human review.
Common mistakes
- Treating f⁻¹(x) as 1/f(x).
- Reversing a many-to-one mapping and silently choosing one of its original inputs.
- Selecting a square-root sign without using the original domain.
- Extending the inverse to a stated target value outside the actual original image.
- Restoring a cancelled original hole or an excluded practical endpoint.
- Keeping the original input/output units in their original positions after reversing the mapping.
What can you explain now?
Does x² on the whole interval [−2,2] have an inverse function merely because output0has only one original input?
Compare with the explanation
No.
Other original outputs,including1and4,each have two allowed original inputs. The entire reversed relation is not a function.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Close the reference. Retain the original domain,identify its actual image,establish uniqueness or give a collision,derive the appropriate inverse branches,then check recovery on the allowed sets and reverse the units.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.