MATH 12002 - CALCULUS I 5.1: Calculus of Inverse Functions - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 5.1: Calculus of Inverse Functions Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 9

Inverse Functions Review algebra of inverse functions in the text, including: one-to-one functions and the horizontal line test; definition of inverse function, including domain and range; cancellation property (or inverse function property); finding the inverse of a one-to-one function; graph of the inverse of a function. D.L. White (Kent State University) 2 / 9

Inverse Functions Definition A function f with domain D is one-to-one if whenever x 1 � = x 2 in D , we have f ( x 1 ) � = f ( x 2 ). Definition If f is a one-to-one function with domain D and range R , then the inverse of f is the function f − 1 with domain R and range D , defined by f − 1 ( b ) = a ⇐ ⇒ f ( a ) = b . D.L. White (Kent State University) 3 / 9

Inverse Functions Graph of an Inverse If y = f ( x ) is a one-to-one function, then the graph of f − 1 is the reflection in the line y = x of the graph of f . Cancellation (Inverse Function Property) If f is a one-to-one function with domain D and range R, then f − 1 ( f ( x )) = x for all x in D and f ( f − 1 ( x )) = x for all x in R . D.L. White (Kent State University) 4 / 9

Continuity and Differentiability Recall our graphical interpretations of continuity and differentiability: A function f is continuous on an interval if and only if the graph of f has no holes, breaks, or jumps on the interval. A continuous function f is differentiable at x = a if and only if the graph of f does not have a “sharp corner” at x = a , and does not have a vertical tangent line at x = a . That is, f is differentiable if its graph is “smooth” and does not have a vertical tangent line. Observe that reflecting a graph across the line y = x cannot introduce a hole, break, jump, or corner in the graph. However, a vertical tangent for the reflected graph will occur at any point where the original graph has a horizontal tangent. Therefore, we have the following result: D.L. White (Kent State University) 5 / 9

Continuity and Differentiability Theorem Let y = f ( x ) be a one-to-one function. If f is continuous at a and f ( a ) = b, then f − 1 is continuous at b. If f is differentiable at a and f ( a ) = b, then f − 1 is differentiable at b unless f ′ ( a ) = 0 . How is the derivative of f − 1 related to f and its derivative? Recalling that f ( f − 1 ( x )) = x and taking derivatives of both sides, we have f ′ ( f − 1 ( x )) · ( f − 1 ) ′ ( x ) = 1 . Hence 1 ( f − 1 ) ′ ( x ) = f ′ ( f − 1 ( x )) for any x such that f ′ ( f − 1 ( x )) � = 0. D.L. White (Kent State University) 6 / 9

Examples 1 Find ( f − 1 ) ′ (10) for f ( x ) = x 3 + 2. By the formula, we have 1 ( f − 1 ) ′ (10) = f ′ ( f − 1 (10)) . We know that f ′ ( x ) = 3 x 2 . Also, f − 1 (10) is the number x such that f ( x ) = 10, √ that is, x 3 + 2 = 10, so x 3 = 8 and x = 3 8 = 2. Hence f − 1 (10) = 2 , and so 1 1 3(2 2 ) = 1 1 ( f − 1 ) ′ (10) = f ′ ( f − 1 (10)) = f ′ (2) = 12 . [Continued → ] D.L. White (Kent State University) 7 / 9

Examples [Example 1, continued] In this example, we could also find f − 1 ( x ) explicitly: x 3 + 2 y = x 3 = y − 2 � 3 x = y − 2 , √ x − 2. and interchanging x and y we have f − 1 ( x ) = y = 3 Therefore, ( f − 1 ) ′ ( x ) = 1 1 3( x − 2) − 2 / 3 = √ x − 2) 2 , 3( 3 and so 1 1 3(2 2 ) = 1 1 ( f − 1 ) ′ (10) = √ 10 − 2) 2 = √ 8) 2 = 12 , 3( 3 3 3( confirming our previous result. D.L. White (Kent State University) 8 / 9

Examples dx x n = nx n − 1 , d Recall that we proved the power rule for derivatives, only for a positive integer n . 2 Prove that if g ( x ) = x 1 / n for a positive integer n , then g ′ ( x ) = 1 1 n − 1 . nx √ x is the inverse of the function f ( x ) = x n The function g ( x ) = x 1 / n = n (with range [0 , ∞ ) if n is even). Hence f ( g ( x )) = x for x in the domain of g . We have already shown that f ′ ( x ) = nx n − 1 since n is a positive integer. Therefore, taking derivatives of both sides of the equation f ( g ( x )) = x , we obtain f ′ ( g ( x )) g ′ ( x ) = 1, and so 1 1 1 1 = 1 1 n − 1 . g ′ ( x ) = f ′ ( g ( x )) = n ( g ( x )) n − 1 = n ( x 1 / n ) n − 1 = nx nx 1 − 1 n D.L. White (Kent State University) 9 / 9