IB Mathematics AA HL Flashcards- Composite/Inverse functions

COMPOSITE FUNCTIONS ➔ Instead of just plugging a number into \(x\) in a function, a composite function will plug in a whole other function into \(x\), creating a new function of \(x\).

[q]

NOTATION ➔ \((f \cdot g)(x)\) means that you are inserting \(g(x)\) into \(f(x)\). In other words:

\[
(f \cdot g)(x) = f(g(x))
\]

You can either view this as one, new, function. Or you may see it as a two-step process that runs \(x\) through the ‘g’ function, then runs that result through the ‘f’ function.

[a]

E.G.  ➔ If \(f(x) = 2x + 1\) & \(g(x) = 3 – 4x\), find \((f \cdot g)(x)\):

\[
f(g(x)) = f(3 – 4x) = 2(3 – 4x) + 1 = 6 – 8x + 1 = 7 – 8x
\]

[q]

We can also evaluate a composite function using a value of \(x\):

E.G. 2 ➔ If \(f(x) = 6x – 5\) & \(g(x) = x^2 – x\), find \((g \cdot f)(1)\):

Either find \((g \cdot f)(x)\) and plug in 1 **or** do \(f(1)\), then \(g(f(1))\):

1. \((g \cdot f)(x) = (6x – 5)^2 – (6x – 5) = 36x^2 – 60x + 25 – 6x + 5 = 36(1)^2 – 60(1) + 25 – 6(1) + 5 = 0\)
2. \(f(1) = 1\), \(g(f(1)) = g(1) = (1)^2 – 1 = 0\)

[a]
INVERSE ➔ Finding the inverse function has been covered in 2.2, but inverse functions are often combined with composites:E.G.  ➔ If \(g(x) = \sqrt{x+1}\) & \(h(x) = 5x – 3\), find \((h \cdot g^{-1})(x)\):

Firstly, find \(g^{-1}(x)\):
\(y = \sqrt{x+1} \Rightarrow x = y^2 + 1 \Rightarrow g^{-1}(x) = x^2 – 1\)

Then, \(h(g^{-1}(x)) = 5(x^2 – 1) – 3 = 5x^2 – 5 – 3 = 5x^2 – 8\)

[q]
CHECKING INVERSES ➔ More significantly, given the nature of inverse functions, we get the following result: If you run a number through a function, then plug the result into the inverse function, you get back to the original number, by definition, i.e.:

\[
(f \cdot f^{-1})(x) = x
\]

[a]
E.G. ➔ Are \(f(x) = 2(\frac{x^2 – 3}{x + 3})\) & \(g(x) = \log_3 (\frac{x+2}{3}) + 1\) inverses of each other?

Check \((f \cdot g)(x)\):

\[
2 \left( 4 \log_3 \left( \frac{x+2}{3} \right) + 1 – 1 \right) – 3 = 2 \log_3 \left( \frac{x+3}{2} \right) – 3 = x + 3 – 3 = x
\]

[x] Exit text