2. 1st example, begin with your function f(x) = 3x – 7 replace f(x) with y y = 3x - 7 Interchange x and y to find the inverse x = 3y – 7 now solve for y x + 7 = 3y = y f-1(x) = replace y with f-1(x) Finding the inverse
3. 2nd example g(x) = 2x3 + 1 replace g(x) with y y = 2x3 + 1 Interchange x and y to find the inverse x = 2y3 + 1 now solve for y x - 1 = 2y3 = y3 = y g-1(x) = replace y with g-1(x) Finding the inverse
4. Recall, to verify you have found the inverse you check that composition of the function with the inverse, in both orders, equals x Using specific ordered pairs can illustrate how the inverse works, but does not verify that it is the inverse. Verifying inverses
5. Consider f(x) = What is the domain? x + 4 > 0 x > -4 or the interval [-4, ∞) What is the range? y > 0 or the interval [0, ∞) Function with a restricted domain
6. Now find the inverse: f(x) = D: [-4, ∞) R: [0, ∞) y = Interchange x and y x = x2 = y + 4 x2 – 4 = y f-1(x) = x2 – 4 D: [0, ∞) R: [-4, ∞) Function with a restricted domain
7. Finally, let us consider the graphs: f(x) = D: [-4, ∞) R: [0, ∞) blue graph f-1(x) = x2 – 4 D: [0, ∞) R: [-4, ∞) red graph Functions with a restricted domain
8. 2nd example Consider g(x) = 5 - x2 D: [0, ∞) What is the range? Make a very quick sketch of the graph R: (-∞, 5] Function with a restricted domain
9. Now find the inverse: g(x) = 5 - x2 D: [0, ∞) R: (-∞, 5] y = 5 - x2 Interchange x and y x = 5 - y2 x – 5 = -y2 5 – x = y2 = y but do we want the + or – square root? g-1(x) = D: (-∞, 5] R: [0, ∞) Function with a restricted domain
10. And, now the graphs: g(x) = 5 - x2 D: [0, ∞) R: (-∞, 5] blue graph g-1(x) = D: (-∞, 5] R: [0, ∞) red graph Functions with a restricted domain
11. A function is one-to-one if each x and y-value is unique Algebraically it means if f(a)=f(b), then a=b. On a graph it means the graph passes the vertical and the horizontal line tests. If a function is one-to-one it has an inverse function. One-to-one