Composite and inverse functions can be determined for trigonometric, logarithmic, exponential or algebraic functions.
Part ofMathsAlgebraic and trigonometric skills
If you're feeling confident, try this more difficult example. Don't be worried if it looks hard, the process you go through is just the same as before. Replace \(x\) with the function and then simplify.
\(f(x) = \frac{{x - 1}}{{x + 1}}\)
Find \(h(x) = f(f(x))\)
\(f(f(x))=\frac{f(x)-1}{f(x)+1}\)
\( = \frac{{\frac{{x - 1}}{{x + 1}} - 1}}{{\frac{{x - 1}}{{x + 1}} + 1}}\)
You now need to simplify the numerator and denominator separately.
Numerator
\(\frac{{x - 1}}{{x + 1}} - 1\)
\(= \frac{{x - 1}}{{x + 1}} - \frac{{x + 1}}{{x + 1}}\)
\(= \frac{{x - 1 - (x + 1)}}{{x + 1}}\)
\(= \frac{{ - 2}}{{x + 1}}\)
Denominator
\(\frac{{x - 1}}{{x + 1}} + 1\)
\(=\frac{{x - 1}}{{x + 1}} + \frac{{x + 1}}{{x + 1}}\)
\(= \frac{{x - 1 + x + 1}}{{x + 1}}\)
\(= \frac{{2x}}{{x + 1}}\)
Now we have:
\(f(f(x))=\frac{\frac{-2}{x+1}}{\frac{2x}{x+1}}\)
\(= \frac{{ - 2}}{{x + 1}} \div \frac{{2x}}{{x + 1}}\)
\(= \frac{{ - 2}}{{x + 1}} \times \frac{{x + 1}}{{2x}}\)
\(= \frac{{ - 2(x + 1)}}{{2x(x + 1)}}\)
\(= \frac{{ - 2}}{{2x}}\)
\(= \frac{{ - 1}}{x}\)
Therefore:
\(h(x) = \frac{{ - 1}}{x}\)
