Question
How to find a composition and inverse when the function has a restriction
Original question: The MS is done by MadasMaths so it's not an actual question from a paper. I do think he made a mistake but it's not that important, what is important is that you understand to take the domain from the inside function and the range of the inside function as domain of the outside function. Here is a similar question I gave a couple of years ago in an end of term progress test: Question 13 g(x)=\frac{2x+5}{x-3}, x\ge 5 (a) Find gg(5). (b) Find g^{-1}(x), stating its domain. (c) State the range of gg(x).
Expert Verified Solution
Key takeaway: These questions look messy only because there are two things happening at once: composition and inverse functions. Once you separate the steps, the domain and range bookkeeping becomes much more manageable.
For a function like you have to respect the restriction at every stage.
(a) Find
First evaluate the inside function: Now plug that into :
=\frac{15+5}{\frac{9}{2}}=\frac{20}{9/2}=\frac{40}{9}$$ So, $$g(g(5))=\frac{40}{9}.$$ ### (b) Find $g^{-1}(x)$ Start with $$y=\frac{2x+5}{x-3}$$ Swap $x$ and $y$: $$x=\frac{2y+5}{y-3}$$ Solve for $y$: $$x(y-3)=2y+5$$ $$xy-3x=2y+5$$ $$xy-2y=3x+5$$ $$y(x-2)=3x+5$$ $$y=\frac{3x+5}{x-2}$$ So $$g^{-1}(x)=\frac{3x+5}{x-2}.$$ Now the domain of the inverse is the range of $g$ on $x\ge 5$. Check the endpoint and behavior: $$g(5)=\frac{15}{2}$$ and for $x\ge 5$, the function decreases toward $2$ but never reaches it. So the range is $$(2,\tfrac{15}{2}].$$ That becomes the domain of $g^{-1}$. ### (c) State the range of $g(x)$ a> For $x\ge 5$, the range is $$(2,\tfrac{15}{2}].$$ --- **Pitfalls the pros know** 👇 The main trap is forgetting the restriction $x\ge 5$. You cannot treat the function as if it were defined for all real numbers. Another common slip is writing the inverse correctly but giving the wrong domain for it. The inverse’s domain must come from the original function’s range, not from the algebra alone. **What if the problem changes?** If the question instead asked for $g(g(x))$, you would first write the composition symbolically: $$g(g(x))=g\left(\frac{2x+5}{x-3}\right),$$ then simplify carefully and keep track of where each denominator is zero. If the restriction were different, say $x\ge a$, the range of $g$ could change, and that would change the domain of $g^{-1}$. The inverse formula might stay the same, but the allowed inputs would not. `Tags`: function composition, inverse function, domain and rangeFAQ
How do I find the inverse of a rational function with a domain restriction?
Solve y=f(x) for x, then swap variables to get the inverse. After that, use the original function’s range as the inverse’s domain, especially when the original function has a restriction.
Why does the domain restriction matter in function composition?
Because the inside function must produce an input that is allowed for the outside function. If the range of the inside function does not fit the outside function’s domain, the composition is not valid for those inputs.