1. EXAMPLE 1 Find an inverse relation Find an equation for the inverse of the relation y = 3x – 5. Write original relation. y = 3x – 5 Switch x and y. x = 3y – 5 Add 5 to each side. x + 5 = 3y Solve for y. This is the inverse relation. 1 3 x +x + 5 3 = y= y

2. EXAMPLE 2 Verify that functions are inverses Verify that f(x) = 3x – 5 and f –1 (x) = 1 3 x + 5 3 are inverse functions. STEP 1 Show: that f(f –1 (x)) = x. f (f –1 (x)) = f 3 1 x + 5 3 = x + 5 – 5 = x SOLUTION 3 1 x + 5 3 = 3 – 5 STEP 2 Show: that f –1 (f(x)) = x. = 1 3 5 3 (3x – 5) + = x – 5 3 5 3 + = x f –1 (f(x)) = f –1 ((3x – 5)

3. EXAMPLE 1 Graph a square root function Graph y =,and state the domain and range. Compare the graph with the graph of y =. 1 2  x  x SOLUTION Make a table of values and sketch the graph.

4. EXAMPLE 1 Graph a square root function The radicand of a square root must be nonnegative. So, the domain is x ≥ 0. The range is y ≥ 0. The graph of y = is a vertical shrink of the graph of y = by a factor of.  x 1 2  x 1 2

5. EXAMPLE 2 Graph a cube root function Graph y = –3 3, and state the domain and range. Compare the graph with the graph of y =.  x x  3 SOLUTION Make a table of values and sketch the graph.

6. EXAMPLE 2 Graph a cube root function The domain and range are all real numbers. The graph of y = – 3 is a vertical stretch of the graph of y = by a factor of 3 followed by a reflection in the x- axis. x  3 x  3

7. EXAMPLE 4 Graph a translated square root function Graph y = – 2 x – 3 + 2. Then state the domain and range. SOLUTION STEP 1 Sketch the graph of y = – 2 x (shown in blue). Notice that it begins at the origin and passes through the point (1, – 2).

8. EXAMPLE 4 Graph a translated square root function STEP 2 Translate the graph. For y = – 2 x – 3 + 2, h = 3 and k = 2. So, shift the graph of y = – 2 x right 3 units and up 2 units. The resulting graph starts at (3, 2) and passes through (4, 0). From the graph, you can see that the domain of the function is x ≥ 3 and the range of the function is y ≤ 2.

9. EXAMPLE 5 Graph a translated cube root function Graph y = 3 3 x + 4 – 1. Then state the domain and range. SOLUTION STEP 1 Sketch the graph of y = 3 3 x ( shown in blue ). Notice that it begins at the origin and passes through the point (– 1, – 3) and (1, 3).

10. EXAMPLE 5 Graph a translated cube root function STEP 2 Translate the graph. Note the for y = – 3 3 x + 4 – 1, h = – 4 and k = – 1. So, shift the graph of y = 3 3 x left 4 units and up 1 unit. The resulting graph starts at (– 5, – 4),(– 4, – 1) and passes through (– 3, 2). From the graph, you can see that the domain and range of the function are both all real numbers.

11. EXAMPLE 1 Solve a radical equation Solve 3 = 3.  2x+7  3 = 3 = 3 3  2x+7 3 ( ) 3 2x+7 = 27 2x2x= 20 x = 10 Write original equation. Cube each side to eliminate the radical. Simplify. Subtract 7 from each side. Divide each side by 2.

12. EXAMPLE 1 Solve a radical equation CHECK Check x = 10 in the original equation. Substitute 10 for x. Simplify. Solution checks. = 3 3  2(10)+7 3 3 = ? 27  3 3 = ?

13. GUIDED PRACTICE for Example 1 Solve equation. Check your solution. 1. 3 √ x – 9 = –1 x = 512 Write original equation. Add 9 to each side. Use each side to eliminate the radical. 3 √ x – 9 = –1 3 √ x = 8 ( 3 √ x ) 3 = (8) 3

14. Standardized Test Practice EXAMPLE 3 SOLUTION 4x 2/3 = 36 Write original equation. x 2/3 = 9 Divide each side by 4. (x 2/3 ) 3/2 = 9 3/2 x = 27 Simplify. The correct answer is C. ANSWER Raise each side to the power. 3 2

15. Solve an equation with a rational exponent EXAMPLE 4 Solve (x + 2) 3/4 – 1 = 7. (x + 2) 3/4 – 1 = 7 (x + 2) 3/4 = 8 (x + 2) 3/4 4/3 = 8 4/3 x + 2 = (8 1/3 ) 4 x + 2 = 2 4 x + 2 = 16 Write original equation. Add 1 to each side. Simplify. Apply properties of exponents. Raise each side to the power. 4 3 Simplify. x = 14 Subtract 2 from each side.

16. Solve an equation with a rational exponent EXAMPLE 4 The solution is 14. Check this in the original equation. ANSWER

17. Solve an equation with an extraneous solution EXAMPLE 5  Solve x + 1 = 7x + 15.  x + 1 = 7x + 15 (x + 1) 2  = ( 7x + 15) 2 x 2 + 2x + 1 = 7x + 15 x 2 – 5x – 14 = 0 (x – 7)(x + 2) = 0 x – 7 = 0 or x + 2 = 0 Write original equation. Square each side. Expand left side and simplify right side. Write in standard form. Factor. Zero - product property x = 7 or x = –2 Solve for x.

18. Solve an equation with an extraneous solution EXAMPLE 5 CHECK 1 x + 1 = 7x + 15  7 + 1 ?  = 7(7) + 15 8 = 64  ? 8 = 8 x + 1 = 7x + 15  –2 + 1 ?  = 7(–2) + 15 –1 = 1  ? /–1 Check x = 7 in the original equation. Check x = –2 in the original equation. The only solution is 7. (The apparent solution 22 is extraneous.) ANSWER

19. Solve an equation with two radicals EXAMPLE 6 x + 2 + 1   = 3 – x x + 2 + 1  2  = 3 – x 2  x + 2 +1 x + 2 +2 = 3 – x  2 x + 2 = – 2x SOLUTION METHOD 1 Solve using algebra. Write original equation. Square each side. Expand left side and simplify right side. Isolate radical expression. x + 2 + 1   = 3 – x. Solve

20. Solve an equation with two radicals EXAMPLE 6 x + 2  = –x  x + 2 2 = ( –x) 2 = x 2 0= x 2 – x – 2 = (x – 2)(x + 1) 0 x – 2 = 0x + 1 = 0 or x = 2 or x = –1 Divide each side by 2. Square each side again. Simplify. Write in standard form. Factor. Zero-product property. Solve for x.