Section 1.6 – Inverse Functions Section 1.7 - Logarithms

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Transcript of Section 1.6 – Inverse Functions Section 1.7 - Logarithms

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Section 1.6 Inverse Functions Section 1.7 - Logarithms Slide 2 For an inverse to exist, the function must be one-to-one (For every x is there is no more than one y) (For every y is there is no more than one x) (Must pass BOTH vertical and horizontal line test) (No x value repeatsno y value repeats) As with all AP Topics, we will look at it: Numerically Graphically Analytically Using a methodology involving algebra or other methods Slide 3 NUMERICALLY Since no x-value has more than one y-value it IS a function. Since no y-value has more than one x-value it IS one-to-one THEREFORE, an inverse exists. represents the inverse of the function Domain of the inverse function is 2.2, 2.02, 2, 1.98, 1.8 Range of the inverse function is 1.1, 1.01, 1, 0.99, 0.9 Slide 4 GRAPHICALLY EXAMPLE #1 f(x) Slide 5 GRAPHICALLY EXAMPLE #2 f(x) (-2, 2) (0, 3) (1, 1) (3, -1) (-2, -2) (-3, -3) Slide 6 ANALYTICALLY 1.Since f(x) represents a line, it is one-to-one, and thus an inverse exists. 2. To find the inverse, switch the x and y, and re-solve for y Slide 7 Slide 8 1. Is g(x) one-to-one????? Note: The inverse function is suddenly NOT one-to-one Slide 9 WARNING WARNING WARNING The original function is NOT one-to-one, so no inverse exists. Slide 10 Evaluate: Solve for x: Slide 11 Simplify: Slide 12 X Slide 13 X X Slide 14