E2.13 - Functions, domain and range

E2.13 - Functions, domain and range

A function is a rule with specified allowed inputs, not just an expression on its own. You will use function notation and mappings, decide which inputs are permitted and find the outputs a function can actually produce. This lesson is for Extended.

Function notation and mappings

Can two different inputs give the same output without breaking a function? A machine that squares its input sends both −2-2 and 2 to 4. That is allowed: each input still has exactly one output. Sending the single input 2 to both 4 and 7 would be different and would not define a function.

Function and notation

A function assigns exactly one output to each input in its domain, the set of allowed inputs. The range is the set of outputs actually obtained. In f(x)=2x+1f(x)=2x+1, ff names the function and f(x)f(x), read “f of x”, names its output at input xx. The brackets do not mean multiplication by ff.

A mapping diagram lists allowed inputs on the left and outputs on the right, with one arrow from each input. Several arrows may arrive at the same output. Unused values in a larger right-hand set are not part of the range: only values reached by arrows belong to it.

[DIAGRAM: asset_name: Linear and many-to-one mappings; asset_slug: function_mappings; caption: Several inputs may share an output, but each input has just one arrow.; description: Two mapping diagrams use the domain negative two, zero, two. The left is labelled f of x equals two x plus one and maps to negative three, one, five. The right is labelled g of x equals x squared plus one and sends both negative two and two to five and zero to one. The right-hand range contains one and five only. All arrows have clear endpoints.]
Several inputs may share an output, but each input has just one arrow.

Worked example 1: evaluating a positive input

Question. For f(x)=2x+1f(x)=2x+1, find f(3)f(3).

Plan. Replace every occurrence of the input letter with 3, then follow the operations.

f(3)=2(3)+1substitute the input=6+1=7multiply before adding\begin{aligned} f(3)&=2(3)+1 && \text{substitute the input}\\ &=6+1=7 && \text{multiply before adding} \end{aligned}

Answer. f(3)=7f(3)=7.

Check. Undo the two operations on 7: subtract 1 and divide by 2 to recover 3. This confirms the machine's input–output pair.

Change only the sign of the input. Brackets keep that sign attached to the substituted value.

Worked example 2: evaluating a negative input

Question. For the same function, find f(−3)f(-3).

Plan. Substitute the whole signed input into the same rule.

f(−3)=2(−3)+1input is negative three=−6+1=−5use signed arithmetic\begin{aligned} f(-3)&=2(-3)+1 && \text{input is negative three}\\ &=-6+1=-5 && \text{use signed arithmetic} \end{aligned}

Answer. f(−3)=−5f(-3)=-5.

Check. (−5−1)/2=−3(-5-1)/2=-3, so reversing the arithmetic recovers the new input. Also, moving the input down by 6 changes the output by 2(−6)=−122(-6)=-12: 7−12=−57-12=-5.

If a domain is finite, evaluate each allowed input. For g(x)=x2+1g(x)=x^2+1 on {−2,0,2}\{-2,0,2\}, the outputs are 5,1,55,1,5, so the range is {1,5}\{1,5\}. Sets do not list repeated values twice. The domain has three members and the range only two; that does not stop gg being a function.

Watch out

f(3)f(3) is an output, while “solve f(x)=3f(x)=3” asks for an input. In the latter, solve 2x+1=32x+1=3, giving x=1x=1; do not simply substitute 3.

Use this different rule and finite domain.

Domain restrictions

The written rule may forbid some real inputs, and a question may deliberately permit fewer inputs still. For example, 1/(x−4)1/(x-4) cannot accept 4 because division by zero has no value. A model for time since a start may restrict an otherwise valid algebraic rule to x≥0x\ge0.

Finding the real domain

If no smaller domain is stated, use all real inputs for which the rule is defined. A denominator must be non-zero. A square root needs a non-negative expression inside it; if that root is itself a denominator, it must be strictly positive. Apply all restrictions together.

The square-root convention matters: 9=3\sqrt9=3, a single non-negative output, rather than ±3\pm3. The equation u2=9u^2=9 has two solutions, but that is a different statement. A function must assign one output to each permitted input.

Worked example 3: a shifted square root

Question. State the largest real domain of p(x)=x−4p(x)=\sqrt{x-4}.

Plan. Require the expression inside the root to be non-negative.

x−4≥0real square root conditionx≥4add four\begin{aligned} x-4&\ge0 && \text{real square root condition}\\ x&\ge4 && \text{add four} \end{aligned}

Answer. All real x≥4x\ge4.

Check. At 4, the output is 0=0\sqrt0=0, so equality is allowed. At 3, the rule would require −1\sqrt{-1}, which has no real value.

Keep the same root but put it in a denominator. This one change now excludes the endpoint too.

Worked example 4: the root becomes a denominator

Question. State the largest real domain of q(x)=1/x−4q(x)=1/\sqrt{x-4}.

Plan. Combine the real-root condition with the non-zero denominator condition.

x−4≥0root must be realx−4≠0division by zero is forbiddenx−4>0⇒x>4combine the conditions\begin{aligned} x-4&\ge0 && \text{root must be real}\\ \sqrt{x-4}&\ne0 && \text{division by zero is forbidden}\\ x-4&>0\Rightarrow x>4 && \text{combine the conditions} \end{aligned}

Answer. All real x>4x>4.

Check. At 5, the output is 1, so a permitted input works. At 4, the expression becomes 1/01/0, confirming the strict boundary. Inputs below 4 still fail the real-root condition.

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