Inverse Functions and Logarithms

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1 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka Inverse Functions and Logarithms DEFINITION: A function f is called a one-to-one function if it never takes on the same value twice; that is, f(x 1 ) f(x ) whenever x 1 x one-to-one not one-to-one HORIZONTAL LINE TEST: A function is one-to-one 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 one-to-one, since if x 1 x, then x 1 x, x 3 1 x 3, x 5 1 x 5, 1 x 1 1 x. Functions x, x, sinx, etc. are not one-to-one, since ( 1) = 1, ( 1) = 1, sin = sinπ DEFINITION: Let f be a one-to-one function with domain A and range B. Then its inverse function f 1 has domain B and range A and is defined by f 1 (y) = x f(x) = y ( ) for any y in B. So, we can reformulate ( ) as (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. 1

2 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka (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 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 ( ) ( ) 1 1 x = x = x. 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) = 7x+, then f 1 (x) = x, since 7 f 1 (f(x)) = f 1 (7x+) = (7x+) 7 ( ) ( x x = x and f(f 1 (x)) = f = ) + = x Step : Solve for x: y = 7x+ y = 7x+ = y = 7x = y 7 = x 6. Let f(x) = (3x ) 5 +. Find f 1 (x). f 1 (x) = x 7

3 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka 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 7. Let f(x) = 3x 5 x. Find f 1 (x). f 1 (x) = 5 x Let f(x) = x. Find f 1 (x). 3

4 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka 7. Let f(x) = 3x 5 x, then f 1 (x) = x+5 3+x. Step : Solve for x: y = 3x 5 x y = 3x 5 x = y( x) = 3x 5 = y xy = 3x 5 = y+5 = 3x+xy y +5 = x(3+y) = y +5 3+y = x 8. Let f(x) = x, then f 1 (x) = x, x. IMPORTANT: f 1 (x) = x+5 3+x domain of f 1 = range of f range of f 1 = domain of f 9. Let f(x) = 3x, then f 1 (x) = 1 3 (x +), x (see Appendix, page 9). 1. Let f(x) = x 1, then f 1 (x) = x +1, x (see Appendix, page 9). 11. Let f(x) = x+5+1, then f 1 (x) = (x 1) 5, x 1 (see Appendix, page 1). 1. Let f(x) = x 7+5. Find f 1 (x).

5 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka domain of f 1 = range of f range of f 1 = domain of f 1. Let f(x) = x 7+5, then f 1 (x) = (x 5) +7, x 5 (see Appendix, page 1). 13. The function f(x) = x is not invertible, since it is not a one-to-one function. REMARK: Similarly, are not invertable functions. x, x 1, sinx, cosx, etc. 1. The function f(x) = (x+1) is not invertible. 15. Let f(x) = x,x, then f 1 (x) = x,x. 16. Let f(x) = x,x, then f 1 (x) = x,x. 17. Let f(x) = x,x < 3, then f 1 (x) = x,x > The function f(x) = x,x > 1 is not invertible. 19. Let f(x) = (x+1),x > 3. Find f 1 (x).. Let f(x) = (1+x),x 1. Find f 1 (x). 5

6 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka 19. Let f(x) = (x+1),x > 3, then f 1 (x) = x 1,x > 16 (see Appendix, page 11). x+1. Let f(x) = (1+x),x 1, then f 1 (x) =,x 1 (see Appendix, page 11). 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. THEOREM: If f is a one-to-one continuous function defined on an interval, then its inverse function f 1 is also continuous. THEOREM (Differentiability of Inverse Functions): If f is a one-to-one differentiable functionwith inversefunctionf 1 and f (f 1 (a)), thentheinversefunction isdifferentiable at a and (f 1 ) (a) = EXAMPLE: If f(x) = x 5 +x+, find (f 1 ) (). Solution 1: We have (f 1 ) () = 1 f (f 1 (a)) 1 f (f 1 ()). Since f(1) =, it follows that f 1 () = 1. Hence (f 1 ) () = 1 f (f 1 ()) = 1 f (1) But f (x) = 5x +1, f (1) = = 6. This yields (f 1 ) () = 1 f (1) = 1 6 Solution : One can see that y = f 1 (x) satisfies the equation x = y 5 + y +. To find y we differentiate both sides: x = (y 5 +y +) = 1 = 5y y +y = 1 = y (5y +1) = y 1 = 5y +1 Note that if x =, then y = 1 (solution of = y 5 +y +). Therefore (f 1 ) () = y 1 () = = 1 6 6

7 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka Logarithmic Functions If a > and a 1, the exponential function f(x) = a x is either increasing or decreasing and so it is one-to one by the Horizontal Line Test. It has an inverse function f 1 (x), which is called the logarithmic function with base a and is denoted by log a x. We have log a x = y a y = x y y y - - x - - x 6 x BASIC PROPERTIES: f(x) = log a x is a continuous function with domain (, ) and range (, ). Moreover, log a (a x ) = x for every x R, a log a x = x for every x > REMARK: It immediately follows from property 1 that log a a = 1, log a 1 = LAWS OF LOGARITHMS: If x and y are positive numbers, then 1. log a (xy) = log a x+log a y. ( ) x. log a = log y a x log a y. 3. log a (x r ) = rlog a x where r is any real number. EXAMPLES: 1. Use the laws of logarithms to evaluate log 3 7 log 3 1. Solution: We have log 3 7 log 3 1 = log 3 ( 7 1 ) = log 3 7 = log = 3log 3 3 = 3 1 = 3. Use the laws of logarithms to evaluate log 1+log 3 log 9. 7

8 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka. Use the laws of logarithms to evaluate log 1+log 3 log 9. Solution: We have log 1+log 3 log 9 = log (1 3) log 9 = log ( ) = log = log = log = 1 = BASIC CALCULUS PROPERTIES: 1. If a > 1, then lim x log a x = and lim x +log ax =. y. If < a < 1, then lim x log a x = and lim x +log ax =. 6 x Natural Logarithms DEFINITION: The logarithm with base e is called the natural logarithm and has a special notation: log e x = lnx BASIC PROPERTIES: 1. ln(e x ) = x for every x R.. e lnx = x for every x >. REMARK: It immediately follows from property 1 that lne = 1 IMPORTANT FORMULA: For any positive a and b (a,b 1) we have In particular, if a = e, then log b x = log ax log a b log b x = lnx lnb 8

9 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka Appendix 9. Let f(x) = 3x, then f 1 (x) = 1 3 (x +), x. 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. 1. Let f(x) = x 1, then f 1 (x) = x +1, x. Step : Solve for x: y = x 1 y = x 1 = y = x 1 x = y +1 f 1 (x) = x +1 Finally, since the range of f is all nonnegative numbers, it follows that the domain of f 1 is x. 9

10 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka 11. 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) = x 7+5, then f 1 (x) = (x 5) +7, x 5. Step : Solve for x: y = x 7+5 y = x 7+5 = y 5 = x 7 = (y 5) = x 7 = (y 5) +7 = x x = (y 5) +7 f 1 (x) = (x 5) +7 Finally, since the range of f is all numbers 5, it follows that the domain of f 1 is x 5. 1

11 Section 3. Inverse Functions and Logarithms 1 Kiryl Tsishchanka 19. 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) x is positive = x = y 1 y = x+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) = = 16 From this it follows that the domain of f 1 is x > 16. x+1. Let f(x) = (1+x),x 1, then f 1 (x) =,x 1. Step : Solve for x: y = (1+x) y = (1+x) x 1 = y = 1+x = y 1 = x 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. 11

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