Advanced Algebra with Trigonometry 4.1 Inverse Functions

1. Advanced Algebra with Trigonometry 4.1 Inverse Functions

2. Inverse Concepts Stand Up Sit Down Go To Sleep Wake Up 2 Steps Forward, then turn Right Turn Left, then 2 Steps Backward Square, Mulitply, then Add Subtract, Divide, then Square Root

3. 4.1 Inverse Functions Inverse relation: a mapping of output values back to their original input values In plain English: switch the x and y!! (x, y) becomes (y, x)!! Write the inverse relation: {(2, 3), (5, 6), (9, 3)} {(3, 2), (6, 5), (3, 9) Was the original relation a function? What about the inverse? Yes No . . . 3 repeats If both the original relation and the inverse relation are functions, we call the two functions inverse functions. NOTE: f−1is read as f inverse. Two functions f and f −1 are inverses of each other provided that f (f −1(x)) = x and f −1(f (x)) = x. Example] Let f(x) = x + 2. f-1(x) = x – 2 Let x = 5. → f−1(x) → x → f → f(x)→ f−1 → x x → f−1 f → x 5 → 5 + 2 → 7 7 → 7 – 2 → 5 5 → 5 – 2 → 3 3 → 3 + 2 → 5

4. Steps to find inverse: 1. Switch x and y 2. Solve for y Think of the inverse as “undoing” the original function. f(x) = 3x + 1 (linear) Think: multiply by 3; then add 1 y = 3x + 1 Write as y =; then switch x and y. x = 3y + 1 Now, solve for y. 3y + 1 = x This step just makes it easier on the eye. 3y = x – 1 Think: subtract 1; then divide by 3. Rename y as f –1(x)

5. Check both function compositions to insure we found the correct inverse!! f(x) = 3x + 1 f(f –1(x)) = f = 3+ 1 = x – 1 + 1 = x (f –1(f(x)) = f –1(3x + 1) = = = x

6. f(x) = x2 + 2, x ≥ 0 (Note the domain restriction: we are only interested in the right side of the parabola.) y = x2 + 2 Switch x and y. x = y2 + 2 y2 + 2 = x Easier on the eye. Solve for y. y2 = x − 2 y = f−1(x) = Because the original domain was restricted to non-zero reals, you are only interested in the positive portion of the solution.

7. Check both function compositions to insure we found the correct inverse!! f(x) = x2 + 2, x ≥ 0 f−1(x) = f(f –1(x)) = f = = x – 2 + 2 = x (f –1(f(x)) = f –1(x2 + 2) = = = x

8. Try this example: f(x) y Quick check: Let x = 8 f(x) = 7 f –1(7) = 8 x

9. Try this example: f(x) = x3 + 5 Quick check: f(2) = 13 f –1 (13) = 2 y = x3 + 5 x = y3 + 5 y3 + 5 = x y3 = x – 5

10. Is the graph a function???Use the Vertical Line Test!!! Is the graph’s inverse a function??? Use the Horizontal Line Test!!! Remember: The graphs of inverses are reflections of each other in the line y = x. f(x)= x2 OOPS!! The graph’s inverse is NOT a function. f -1(x)

11. Some people can just visualize what an inverse graph looks like using the definition. Others of us need a little assistance. My secret: On the original graph, pick two or three easily identified points. Write them down. Switch the x and y coordinates!!! Plot the new points to give you a start on drawing the inverse. Graph the line y = x. The inverse graph will be symmetric about the line y = x. Consider the function shown in the graph: f(x) I would write down the coordinates of the three points indicated: (−2, −1), (0, 0), and (2, 3). f -1(x) Then I would plot (−1, −2), (0, 0), and (3, 2). Now it is easier to sketch the inverse!!

12. Review: To find the inverse of a relation or function: To test if two relations are functions: Don’t forget about any domain restrictions! Switch x and y. Solve for y. Use Composition to verify that f(g(x)) equals x and g(f(x)) equals x