MATH127 Lecture Notes - Lecture 4: Post-It Note, Binary Logarithm

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To invert; subtract 5, then divide by 2. G(x) = (x-5)/2 g(x) undoes f(x), g(x) = f^(-1) (x) 1/(f(x)) X -> f -> f(x) -> g (which is =f^-1) -> x. How to find an inverse: write y = f(x, solve for x in terms of y. Ex; g(x) = 1/3 |(2x+1: y = g(x) = 1/3 |(2x+1) y= 1/3 |(2x+1, 3y = | 2x+1 (3y)^2 = 2x+1 (3y)^2 -1 = 2x. G(x); d = [ - , infinity), r = [0, infinity) The domain and range of the inverse of g(x) is switched (domain of g(x) = range of g^-1(x)) Because there are 2 options, there is no inverse. A function f: d->r is one-to-one if it never takes the same value twice. 2 different inputs always give 2 different outputs. We define a new function f^-1 ; r -> d by f^-1(y) = x when f(x) =y. The inverse of an exponential is a logarithm (sticky note 2)

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