Section 2.7 OnetoOne Functions and Their Inverses


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1 Section. OnetoOne Functions and Their Inverses OnetoOne Functions HORIZONTAL LINE TEST: A function is onetoone if and only if no horizontal line intersects its graph more than once. EXAMPLES: 1. Functions x, x 3, x 5, 1/x, etc. are onetoone, since if x 1 x, then x 1 x, x 3 1 x 3, x 5 1 x 5, 1 x 1 1 x. The function f(x) = 4x 3 is onetoone. In fact, suppose x 1 and x are real numbers such that f(x 1 ) = f(x ). Then 4x 1 3 = 4x 3 4x 1 = 4x x 1 = x Therefore, f is onetoone. 5 x 3. The function f(x) = +1 is onetoone. In fact, suppose x 1 and x are real numbers such that f(x 1 ) = f(x ). Then 5 x1 +1 = 5 x +1 5 x1 = 5 x 5 x1 = 5 x x 1 = x Therefore, f is onetoone. 4. Functions x, x 4, sinx, etc. are not onetoone, since ( 1) = 1, ( 1) 4 = 1 4, sin0 = sinπ 1
2 DEFINITION: Let f be a onetoone function with domain A and range B. Then its inverse function f 1 has domain B and range A and is defined by for any y in B. So, we can reformulate ( ) as f 1 (y) = x f(x) = y (f 1 f)(x) = f 1 (f(x)) = x for every x in the domain of f (f f 1 )(x) = f(f 1 (x)) = x for every x in the domain of f 1 IMPORTANT: Do not confuse f 1 with 1 f. EXAMPLES: 1. Let f(x) = x 3, then f 1 (x) = 3 x, since f 1 (f(x)) = f 1 (x 3 ) = 3 x 3 = x and f(f 1 (x)) = f( 3 x) = ( 3 x) 3 = x. Let f(x) = x 3 +1, then f 1 (x) = 3 x 1, since f 1 (f(x)) = f 1 (x 3 +1) = 3 (x 3 +1) 1 = x and f(f 1 (x)) = f( 3 x 1) = ( 3 x 1) 3 +1 = x ( ) 3. Let f(x) = x, then f 1 (x) = 1 x, since f 1 (f(x)) = f 1 (x) = 1 (x) = x and f(f 1 (x)) = f 4. Let f(x) = x, then f 1 (x) = x, since f 1 (f(x)) = f 1 (x) = x and f(f 1 (x)) = f(x) = x 5. Let f(x) = x+, then f 1 (x) = x, since f 1 (f(x)) = f 1 (x+) = (x+) ( ) ( ) 1 1 x = x = x ( ) ( x x = x and f(f 1 (x)) = f = ) + = x Step : Solve for x: y = x+ y = x+ = y = x = y f 1 (x) = x = x 6. Let f(x) = (3x ) 5 +. Find f 1 (x).
3 6. Let f(x) = (3x ) 5 +. Find f 1 (x). Step : Solve for x: y = (3x ) 5 + y = (3x ) 5 + = y = (3x ) 5 = 5 y = 3x = 5 y + = 3x 5 y + x = 3. Let f(x) = 3x 5 4 x. Find f 1 (x). f 1 (x) = 5 x Let f(x) = x. Find f 1 (x). 3
4 . Let f(x) = 3x 5 4 x, then f 1 (x) = 4x+5 3+x. Step : Solve for x: y = 3x 5 4 x y = 3x 5 4 x = y(4 x) = 3x 5 = 4y xy = 3x 5 = 4y+5 = 3x+xy 4y +5 = x(3+y) = 8. Let f(x) = x, then f 1 (x) = x, x 0. IMPORTANT: f 1 (x) = 4x+5 3+x domain of f 1 = range of f range of f 1 = domain of f 4y +5 3+y = x 9. Let f(x) = 3 x, then f 1 (x) = 3 x, x 0. Step : Solve for x: y = 3 x y = 3 x = y = 3 x = x = 3 y f 1 (x) = 3 x Since the range of f(x) is all nonnegative numbers, it follows that the domain of f 1 (x) is x 0. So, f 1 (x) = 3 x, x 0 4
5 10. Let f(x) = 3x, then f 1 (x) = 1 3 (x +), x 0 (see Appendix, page ). 11. Let f(x) = 4 x 1, then f 1 (x) = x 4 +1, x 0 (see Appendix, page ). 1. Let f(x) = x+5+1, then f 1 (x) = (x 1) 5, x 1 (see Appendix, page 8). 13. Let f(x) = 4 x +5, then f 1 (x) = (x 5)4 +, x 5 (see Appendix, page 8). 14. The function f(x) = x is not invertible, since it is not a onetoone function. REMARK: Similarly, are not invertable functions. x 4, x 10, sinx, cosx, etc. 15. The function f(x) = (x+1) is not invertible. 16. Let f(x) = x,x 0, then f 1 (x) = x,x Let f(x) = x,x, then f 1 (x) = x,x Let f(x) = x,x < 3, then f 1 (x) = x,x > The function f(x) = x,x > 1 is not invertible. 0. Let f(x) = (x+1),x > 3. Find f 1 (x). 1. Let f(x) = (1+x),x 1. Find f 1 (x). 5
6 0. Let f(x) = (x+1),x > 3, then f 1 (x) = x 1,x > 16 (see Appendix, page 9). x+1 1. Let f(x) = (1+x),x 1, then f 1 (x) =,x 1 (see Appendix, page 9). THEOREM: If f has an inverse function f 1, then the graphs of y = f(x) and y = f 1 (x) are reflections of one another about the line y = x; that is, each is the mirror image of the other with respect to that line. EXAMPLE: The graph of a function f is given. Sketch the graph of f 1. 6
7 Appendix 10. Let f(x) = 3x, then f 1 (x) = 1 3 (x +), x 0. Step : Solve for x: y = 3x y = 3x = y = 3x = y + = 3x x = 1 3 (y +) f 1 (x) = 1 3 (x +) Finally, since the range of f is all nonnegative numbers, it follows that the domain of f 1 is x Let f(x) = 4 x 1, then f 1 (x) = x 4 +1, x 0. Step : Solve for x: y = 4 x 1 y = 4 x 1 = y 4 = x 1 x = y 4 +1 f 1 (x) = x 4 +1 Finally, since the range of f is all nonnegative numbers, it follows that the domain of f 1 is x 0.
8 1. Let f(x) = x+5+1, then f 1 (x) = (x 1) 5, x 1. Step : Solve for x: y = x+5+1 y = x+5+1 = y 1 = x+5 = (y 1) = x+5 x = (y 1) 5 f 1 (x) = (x 1) 5 Finally, since the range of f is all numbers 1, it follows that the domain of f 1 is x Let f(x) = 4 x +5, then f 1 (x) = (x 5)4 +, x 5. Step : Solve for x: y = 4 x +5 y = 4 x +5 = y 5 = 4 x = (y 5) 4 = x = (y 5) 4 + = x x = (y 5)4 + f 1 (x) = (x 5)4 + Finally, since the range of f is all numbers 5, it follows that the domain of f 1 is x 5. 8
9 0. Let f(x) = (x+1),x > 3, then f 1 (x) = x 1,x > 16. Step : Solve for x: y = (x+1) y = (x+1) = ± y = x+1 Since x is positive, it follows that y = x+1, x = y 1 f 1 (x) = x 1 To find the domain of f 1 we note that the range of f is all numbers > 16. Indeed, since x > 3, we have f(x) = (x+1) > (3+1) = 4 = 16 From this it follows that the domain of f 1 is x > 16. x+1 1. Let f(x) = (1+x),x 1, then f 1 (x) =,x 1. Step : Solve for x: y = (1+x) y = (1+x) = ± y = 1+x Since x 1, it follows that y = 1+x, hence y +1 y 1 = x = x+1 f 1 (x) = To find the domain of f 1 we note that the range of f is all numbers 1. Indeed, since x 1, we have f(x) = (1+x) (1+ ( 1)) = (1 ) = ( 1) = 1 From this it follows that the domain of f 1 is x 1. = x 9
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