 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/1/f/iv/solution

f (x) = x + 1

and

g (x) = 2 x

used above,

f \circ g

sends

x

2 x + 1

x

(x - 1) /2

, whereas

f^{- 1} \circ g^{- 1}

sends

x

x /2 - 1

. Hence

False in general: (f \circ g)^{- 1} \neq = f^{- 1} \circ g^{- 1} .

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