f(a) is a height; a composite reads twice: To read f(a) off a graph: go up from x = a to the curve, then across to the y-axis.
For (f∘f)(a) do it twice — read f(a) first, then feed that answer back in and read f of it.
IB-style question — read off the graph
The graph of y = f(x) for 0 ≤ x ≤ 8 is shown; it cuts the axes at (0, 6) and (8, 0).
From the graph, f(2) = 3 and f(4) = 2.
Find (a) f(4); (b) (f∘f)(4).
Step by step
- (a) Read the height of the curve at x = 4.
- (b) Composite — inside first: f(4) = 2, then read f(2).
Final answer
(a) f(4) = 2. (b) (f∘f)(4) = 3.
Read f(4): up from x = 4 to the curve, across to y = 2. The composite then reads f(2): up from x = 2 gives y = 3, so (f∘f)(4) = 3.
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Use the answer as the next input: (f∘f)(a) is not f(a) doubled.
Read f(a), then go to that x-value and read the curve again — keep the two read-offs separate.
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f⁻¹ reverses the read — and reflects in y = x: To read f⁻¹(b) off the graph of f: start at y = b on the y-axis, go across to the curve, then down to the x-axis — that x is f⁻¹(b).
To sketch f⁻¹, reflect the whole graph in the line y = x. Every point (a, b) becomes (b, a), so the intercepts swap.
IB-style question — the inverse from the graph
Using the same graph, where f cuts the axes at (0, 6) and (3, 0), and passes through (2, 2):
(a) find f⁻¹(2); (b) state where y = f⁻¹(x) cuts the axes.
Step by step
- (a) f⁻¹(2) asks: 'which input gives output 2?' Across from y = 2, down to the x-axis.
- (b) Reflecting in y = x swaps each intercept's coordinates.
Final answer
(a) f⁻¹(2) = 2. (b) f⁻¹ cuts the axes at (6, 0) and (0, 3).
f⁻¹ is the mirror image of f in the line y = x. Points on the mirror line, like (2, 2), stay in the same place.
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Intercepts swap: A y-intercept (0, k) of f becomes an x-intercept (k, 0) of f⁻¹, and an x-intercept (k, 0) of f becomes a y-intercept (0, k) of f⁻¹.
Mark the reflected intercepts when you sketch.