# Section 1.6 – Inverse Functions Section 1.7 - Logarithms.

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Section 1.6 – Inverse Functions Section 1.7 - Logarithms

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 repeats…no y value repeats) As with all AP Topics, we will look at it: Numerically Graphically Analytically – Using a methodology involving algebra or other methods

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

GRAPHICALLY – EXAMPLE #1 f(x)

GRAPHICALLY – EXAMPLE #2 f(x) (-2, 2) (0, 3) (1, 1) (3, -1) (-2, -2) (-3, -3)

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

1. Is g(x) one-to-one????? Note: The inverse function is suddenly NOT one-to-one

WARNING WARNING WARNING The original function is NOT one-to-one, so no inverse exists.

Evaluate: Solve for x:

Simplify:

X

X X

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