Todays Agenda Upcoming Homework Section 3.2: Inverse Functions and - PowerPoint PPT Presentation

Today’s Agenda • Upcoming Homework • Section 3.2: Inverse Functions and Logarithms Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 1 / 10

Upcoming Homework • WeBWorK HW #12 Section 3.1 due 10/9 • WeBWorK HW #13 Section 3.2 due 10/14 Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 2 / 10

Section 3.2 You’ve no doubt had to compute inverse functions in your previous math classes, but you may not have learned that many functions do not have inverses to begin with! Let’s consider two different functions, f , g : { 1 , 2 , 3 , 4 } → R , defined by the tables below (notice that the domain of both f and g is just four numbers, not an interval): f ( x ) g ( x ) x 1 2 3 2 4 5 3 7 3 4 10 11 What do you notice about the range of f versus the range of g ? Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 3 / 10

Section 3.2 In the tables on the previous slide, we saw that whenever x 1 � = x 2 , f ( x 1 ) � = f ( x 2 ). However, we saw that even though 1 � = 3, g (1) = g (3). There is a special name for the property that the function f possesses, and it is quite important. Definition 3.2.1 A function f : D → R is called an injective function , or a one-to-one function , if it never takes on the same value twice, that is: f ( x 1 ) � = f ( x 2 ) whenever x 1 � = x 2 . A quick way to check if a function is one-to-one is by giving it the ”horizontal line test,” i.e., making sure that any horizontal line intersects the graph of the function at most once. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 4 / 10

Section 3.2 A function f : D → R ⊆ R has an inverse f − 1 : R → D if and only if it is injective. Definition 3.2.2 Let f be a one-to-one function with domain D and range R . Then its inverse function f − 1 has domain R and range D , and is defined by f − 1 ( y ) = x , where f ( x ) = y , for each y ∈ R . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 5 / 10

Section 3.2 Properties of Inverse Functions 1 domain of f − 1 = range of f 2 range of f − 1 = domain of f 3 f − 1 ( f ( x )) = x for all x ∈ D 4 f ( f − 1 ( y )) = y for all y ∈ R Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 6 / 10

Section 3.2 If we are given a function f : D → R ⊆ R and we have determined that f is one-to-one, then to find f − 1 : R → D , we can follow these steps: 1 Write f ( x ) = y . 2 Solve for x in terms of y (if possible). 3 Interchange x and y . 4 After interchanging , write y = f − 1 ( x ). Example 3.2.3 Does the function f : R → R defined by f ( x ) = x 3 + 2 have an inverse? If so, find f − 1 ( x ). A quick note before we continue: If f is continuous, then so is f − 1 . This is actually a theorem in your textbook, which can be found on page 155. Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 7 / 10

Section 3.2 How are the derivatives of inverse functions related to the derivatives of the original functions? We have a theorem for this, which is important enough that is has an actual name. (For some unexplained reason, your textbook doesn’t use the name of the theorem. This is odd because the vast majority of textbooks do use the name, and it’s the same name in all of them.) If a theorem has a name, that usually means it’s a very important one. The Inverse Function Theorem Suppose f : D → R ⊆ R is a one-to-one differentiable function with inverse function f − 1 . Furthermore suppose that for the particular value a ∈ D , f ′ ( f − 1 ( a )) � = 0. Then f − 1 is differentiable at a , and 1 ( f − 1 ) ′ ( a ) = f ′ ( f − 1 ( a )) . Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 8 / 10

Section 3.2 Proof of the Inverse Function Theorem To show that f − 1 is differentiable at a , we need to show that f − 1 ( x ) − f − 1 ( a ) exists. Write f ( b ) = a so that f − 1 ( a ) = b . Since lim x → a x − a f is differentiable, it is continuous, and hence f − 1 is continuous as well. We calculate that f − 1 ( x ) − f − 1 ( a ) y − b 1 lim = lim f ( y ) − f ( b ) = lim x − a f ( y ) − f ( b ) x → a y → b y → b y − b 1 1 1 = = f ′ ( b ) = f ′ ( f − 1 ( a )) . f ( y ) − f ( b ) lim y → b y − b 1 Therefore, f − 1 is differentiable at a and ( f − 1 ) ′ ( a ) = f ′ ( f − 1 ( a )). � Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 9 / 10

Section 3.2 Example Problems 1 Determine whether or not the following functions are one-to-one (injective): f ( t ) is the height of a football t seconds after kickoff. 1 f ( t ) is your height at age t (do not assume old people shrink). 2 2 If f ( x ) = x 5 + x 3 + x , does f − 1 exist? If so, find f − 1 (3) and f ( f − 1 (2)). 3 The inverses of the following functions do exist. Calculate f − 1 ( x ): f ( x ) = 1 + √ 2 + 3 x 1 f ( x ) = e 2 x − 1 2 4 The inverses of the following functions do exist on the given domains, and are differentiable. Find ( f − 1 ) ′ ( a ): f ( x ) = 2 x 3 + 3 x 2 + 7 x + 4, x ∈ R , a = 4 1 f ( x ) = 9 − x 2 , 0 ≤ x ≤ 3, a = 8 2 f ( x ) = 1 / ( x − 1), x > 1, a = 2 3 Lindsey K. Gamard, ASU SoMSS MAT 265: Calculus for Engineers I Wed., 7 October 2015 10 / 10