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A Function of Itself

Part of the Functions & Transformations section of Coddy's Algebra 2 journey. Lesson 20 of 55.

A function can be composed with itself. h(h(x))h\left(h\left(x\right)\right) means: apply h, then apply h again to the output. Take h(x)=3x−1h\left(x\right) = 3x-1 and work out h(h(1))h\left(h\left(1\right)\right). First h(1) is 2. Then h(2) is 5.

Written out, the inner call comes first and gives h(3(1)−1)h\left(3\left(1\right)-1\right), the outer one 3(3(1)−1)−13\left(3\left(1\right)-1\right)-1, and the arithmetic finishes it: 3(2)−13\left(2\right)-1, then 5.

Do not confuse this with squaring or doubling. (h(1))2{\left(h\left(1\right)\right)}^{2} is 4 and 2h(1)2h\left(1\right) is 4, but h(h(1))h\left(h\left(1\right)\right) is 5. Applying a rule twice means running the second pass on the first pass's output, the way a loop feeds each result back in.

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Challenge

For f(x)=2x−3f\left(x\right) = 2x-3, work out f(f(2))f\left(f\left(2\right)\right). Tap Use f(x) = ... twice, the inner call first, then type the arithmetic line by line.

Try it yourself

f(f(2))f\left(f\left(2\right)\right)

Simplify

Givenf(x)=2x−3f\left(x\right) = 2x-3

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