Step By Step Calculus » 4.8 - Inverse of Monotonic Functions

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Synopsis
Monotonic Functions: A function is:
  • Strictly increasing, if for all a, \; ba, \; b in the domain of f , \; \; a< b \; \Rightarrow \; f(a) < f(b) \, . f , \; \; a< b \; \Rightarrow \; f(a) < f(b) \, .
  • Strictly decreasing, if for all a, \; ba, \; b in the domain of f, \; \; a < b \; \Rightarrow \; f(a) > f(b)f, \; \; a < b \; \Rightarrow \; f(a) > f(b).
As an aside, we say ff is monotone increasing if a < b \; \Rightarrow \; f(a) \leq f(b) a < b \; \Rightarrow \; f(a) \leq f(b) , that is we allow equality (the function is either increasing or constant as we move to the right). Dually, we say ff is monotone decreasing if a < b \; \Rightarrow \; f(a) \geq f(b) a < b \; \Rightarrow \; f(a) \geq f(b) .

Inverse of Monotonic Functions: Another way to establish that a function is invertible is by showing that it is strictly increasing or strictly decreasing on its domain: If a function is either strictly increasing, or strictly decreasing, then it is 1-1 and therefore invertible.
Inverse of Composite Functions: The inverse of a composite functionf\circ g(x)f\circ g(x) is (f\circ g)^{-1}(x)=g^{-1}\circ f^{-1}(x).(f\circ g)^{-1}(x)=g^{-1}\circ f^{-1}(x). This leads us to another useful formula \displaystyle \left(\frac{1}{f}\right)^{-1}(x)=f^{-1}\left(\frac{1}{x}\right)\displaystyle \left(\frac{1}{f}\right)^{-1}(x)=f^{-1}\left(\frac{1}{x}\right) when x\neq 0x\neq 0 and \frac{1}{x}\in D_f\frac{1}{x}\in D_f.