2 · IB Maths
Inverse functions
The inverse undoes the function, and its graph is the mirror image in y = x. Drag the point along f and watch its reflection trace out f⁻¹.
Predict first
Choose f(x) = eˣ and set the Point a to 1.0, so the point on f is (1, 2.72). Which point lies on the graph of f⁻¹?
If (a, b) is on f then (b, a) is on f⁻¹, because f⁻¹ is the reflection of f in y = x. Here f⁻¹(x) = ln x, and ln 2.72 = 1.
Check it in the model: Pick f(x) = eˣ in the Function list, set the Point, a slider to 1.0 and read the On f⁻¹ line in the readouts.
f f⁻¹ · drag either point on the graph, or use the slider
f⁻¹(f(a)) = a and f(f⁻¹(x)) = x
method: swap x and y, then rearrange
mirror line: y = x
Try this
- Choose f(x) = 2x + 1 and set a = 1.5. Read off (a, b) and (b, a), then check that f⁻¹(b) gives back a.
- Choose f(x) = eˣ and drag the point. Which line does the graph of f⁻¹ never cross, and how does this match the range of f?
- Choose f(x) = x² and untick the domain restriction. Move the point and explain why the red dot shows that x² has no inverse function.
Exam tip: if (a, b) is on f then (b, a) is on f⁻¹, so the domain and range swap; only a one-to-one function has an inverse, which is why x² needs x ≥ 0 first. To find f⁻¹, swap x and y and rearrange.