Testing a functional equation
A functional equation is checked by applying the stated rule to every complete input. Simplify the left and right sides separately before deciding whether they are equal or finding a missing constant
\[f\left(x+y\right)\]
- Read the complete input The expression inside a function name is one input to the rule, even when that input contains several terms \(\displaystyle f\left(x+y\right)\text{ uses }x+y\text{ as one input}\)
- Substitute before simplifying Replace the function input by the complete expression, then simplify the resulting rule exactly \(\displaystyle f(t)=2t+1\quad\Rightarrow\quad f\left(2+3\right)=2\left(2+3\right)+1=11\)
- Simplify both sides A functional relation is checked by finding the exact value of its left side and its right side separately \(\displaystyle f(2)=5,\quad f(3)=7\quad\Rightarrow\quad f(2)+f(3)=12\)
- Find a missing constant For h(t)=2t+3 and inputs 2 and 1, subtract h(2)+h(1) from h(2+1) to find the additive constant \(\displaystyle C=h(2+1)-h(2)-h(1)=9-12=-3\)
Worked example
For h(t)=2t+3, find the additive constant C in h(2+1)=h(2)+h(1)+C.
\[h(t)=2t+3,\qquad h(2+1)=h(2)+h(1)+C\]
- \(\displaystyle h(2+1)=h(3)=2\left(3\right)+3=9\)Apply the rule to the complete input on the left side
- \(\displaystyle h(2)+h(1)=\left(2\left(2\right)+3\right)+\left(2\left(1\right)+3\right)=7+5=12\)Evaluate the two separate function values on the right side
- \(\displaystyle C=9-12=-3\)Subtract the right-side value from the left-side value to find the additive constant
- \(\displaystyle h(2+1)=h(2)+h(1)-3\)Substitute C=-3 into the functional equation
Common mistakes
Distributing the input
✕\(\displaystyle h\left(2+3\right)=h(2)+h(3)=7+9=16\)✓\(\displaystyle h\left(2+3\right)=2\left(2+3\right)+3=13\)
The function rule must receive the complete input expression before the expression is simplifiedChecking one side only
✕\(\displaystyle h\left(1+2\right)=9\quad\Rightarrow\quad\text{the relation holds}\)✓\(\displaystyle h\left(1+2\right)=9\ne h(1)+h(2)=5+7=12\)
Both exact sides must be evaluated and compared; one side alone cannot establish a functional relationUsing the wrong additive constant
✕\(\displaystyle h(2+1)=h(2)+h(1)+3=15\)✓\(\displaystyle h(2+1)=h(2)+h(1)-3=9\)
The additive constant is the left-side value minus the separate right-side value, so C=9-12=-3.Difficulty levels
introSubstitute complete inputs into a simple rule and compare the two sides
standardSimplify reciprocal or quotient forms exactly on both sides of a relation
challengeFind a missing additive constant or reject a relation using an exact counterexample
examSelect a verdict and reason, then identify exact side values or a missing constant
Questions about this topic
What does f(x+y) mean?+
It means that x+y is substituted as one input into the rule for f. It does not automatically mean f(x)+f(y)
Why must both sides be simplified?+
Equality is established only when the exact value of the left side matches the exact value of the right side
How is a missing additive constant found?+
Calculate the left side and the right side without the constant, then use C=left side−right side
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 3: Functions, transformations and inverses