IB Mathematics AA HL Flashcards- Function and their domain range

A function is a series of operations that will transform any \(x\) value into one specific (y) value.

[q]

NOTATION
The two main versions of writing a function will be: \(y = 4x – 3\) or \(f(x) = 4x – 3\). This takes any \(x\) value, multiplies it by 4, and subtracts 3. For example, \(f(2) = 4(2) – 3 = 5\). But, as usual, you may use any letters for variables, such as a ‘velocity-time’ function like \(v(t) = 3t^2 + 2t\).

[a]

DOMAIN/RANGE
An important detail about functions is knowing what set of \(x\) values you are able to plug in — the domain, and which \(y\) values are possible to get out of the function — the range. The things that may restrict the domain include that: you can’t divide by zero, you ‘can’t do’ the square root of a negative, and you can’t do a log of 0 or a negative. The range essentially has an inverse set of restrictions.

[q]

E.G. 1/4
Find the domain of \(f(x) = \ln(x+1)\):
Must have \(x + 1 > 0\), i.e., \(x > -1\)
Alternate notation: \(\{ x \in \mathbb{R} \ | \ x > -1 \}\)

[a]
INVERSE
We have seen how a regular function turns \(x\)’s into \(y\)’s. The inverse of a function turns those \(y\) values back into their respective \(x\) values, reversing the operations. To find these, we can switch \(x\) & \(y\), then make \(y\) the subject again. Then write it with the notation: \(f^{-1}(x)\) = [inverse function].

[q]

E.G. 4/4
\(h(x) = \ln\left( \frac{x}{2} \right)\), find \(h^{-1}\):

\[
\begin{aligned}
y &= \ln\left( \frac{x}{2} \right) \\
x &= \ln\left( \frac{y}{2} \right) \quad \text{(switch x & y)} \\
e^x &= \frac{y}{2} \quad \text{(raise power of \(e\))} \\
y &= 2e^x
\end{aligned}
\]

Inverse notation: \(h^{-1}(x) = 2e^x\)

[x] Exit text