Presentation is loading. Please wait.

Presentation is loading. Please wait.

Section 1.6 – Inverse Functions Section 1.7 - Logarithms.

Similar presentations


Presentation on theme: "Section 1.6 – Inverse Functions Section 1.7 - Logarithms."— Presentation transcript:

1 Section 1.6 – Inverse Functions Section Logarithms

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 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

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

4 GRAPHICALLY – EXAMPLE #1 f(x)

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

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

7

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

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

10 Evaluate: Solve for x:

11 Simplify:

12 X

13 X X

14


Download ppt "Section 1.6 – Inverse Functions Section 1.7 - Logarithms."

Similar presentations


Ads by Google