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Use the horizontal-line test to determine whether the inverse of the blue relation is a function.

Example 1A: Using the Horizontal-Line Test. Use the horizontal-line test to determine whether the inverse of the blue relation is a function. The inverse is a function because no horizontal line passes through two points on the graph. Example 1B: Using the Horizontal-Line Test.

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Use the horizontal-line test to determine whether the inverse of the blue relation is a function.

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  1. Example 1A: Using the Horizontal-Line Test Use the horizontal-line test to determine whether the inverse of the blue relation is a function. The inverse is a function because no horizontal line passes through two points on the graph.

  2. Example 1B: Using the Horizontal-Line Test Use the horizontal-line test to determine whether the inverse of the red relation is a function. The inverse is a not a function because a horizontal line passes through more than one point on the graph.

  3. Check It Out! Example 1 Use the horizontal-line test to determine whether the inverse of each relation is a function. The inverse is a function because no horizontal line passes through two points on the graph.

  4. Recall from Lesson 7-2 that to write the rule for the inverse of a function, you can exchange x and y and solve the equation for y. Because the value of x and y are switched, the domain of the function will be the range of its inverse and vice versa.

  5. Find the inverse of . Determine whether it is a function, and state its domain and range. Example 2: Writing Rules for inverses Step 1 The horizontal-line test shows that theinverse is a function. Note that the domain and range of f are all real numbers.

  6. Example 2 Continued Step 1 Find the inverse. Rewrite the function using y instead of f(x). Switch x and y in the equation. Cube both sides. Simplify. Isolate y.

  7. Because the inverse is a function, . The domain of the inverse is the range of f(x):{x|xR}. The range is the domain of f(x):{y|yR}. Example 2 Continued CheckGraph both relations to see that they are symmetric about y = x.

  8. Check It Out! Example 2 Find the inverse of f(x) = x3 – 2. Determine whether it is a function, and state its domain and range. Step 1 The horizontal-line test shows that theinverse is a function. Note that the domain and range of f are all real numbers.

  9. 3 3 x + 2 = y 3 3 x + 2 = y Check It Out! Example 2 Continued Step 1 Find the inverse. y= x3 – 2 Rewrite the function using y instead of f(x). x= y3 – 2 Switch x and y in the equation. x + 2= y3 Add 2 to both sides of the equation. Take the cube root of both sides. Simplify.

  10. Because the inverse is a function, . Check It Out! Example 2 Continued The domain of the inverse is the range of f(x): R. The range is the domain of f(x): R. CheckGraph both relations to see that they are symmetric about y = x.

  11. You can use composition of functions to verify that 2 functions are inverses. The inverse functions “undo” each other, When you compose two inverses… the result is the input value of x.

  12. If f(g(x)) = g(f(x)) = x Then f(x) and g(x)are inverse functions Example 1: Because f(g(x)) = g(f(x)) = x, they are inverses.

  13. Substitute x + 1 for x in f. 1 1 3 3 Determine by composition whether each pair of functions are inverses. Example 2: f(x) = 3x – 1 and g(x) = x + 1 Find the composition f(g(x)). Use the Distributive Property. = (x + 3) – 1 f(g(x)) = x + 2 Simplify. The functions are NOT inverses.

  14. f(x) = x + 6 and g(x) = x – 9 3 2 2 3 Determine by composition whether each pair of functions are inverses. Example 3 Find the composition f(g(x)) and g(f(x)). = x – 6 + 6 = x + 9 – 9 Because f(g(x)) = g(f(x)) = x, they are inverses.

  15. Substitute for x in f. = x + 25 +5 - 10 x Example 4 f(x) = x2 + 5 and for x ≥ 0 Find the compositions f(g(x)) and g(f(x)). Simplify. Becausef(g(x)) ≠ x,f (x) and g(x) are NOT inverses.

  16. For x ≠ 1 or 0, f(x) = and g(x) = + 1. 1 1 x x –1 Are the following functions Inverses? Example 5: = (x –1) + 1 Because f(g(x)) = g(f (x)) = xf(x) and g(x) are inverses.

  17. Lesson Quiz: Part I 1. Use the horizontal-line test to determine whether the inverse of each relation is a function. A: yes; B: no

  18. Lesson Quiz: Part II 2. Find the inverse f(x) = x2 – 4.Determine whether it is a function, and state its domain and range. not a function D: {x|x ≥ 4}; R: {all Real Numbers}

  19. Lesson Quiz: Part III 3. Determine by composition whether f(x) = 3(x – 1)2 and g(x) = +1 are inverses for x ≥ 0. yes

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