Transcendental Functions and Inverse Functions in Calculus II Lecture
Understand transcendental functions and inverse functions in Calculus II from this lecture, covering concepts like one-to-one functions, finding inverses, and examples to illustrate the concepts. Gain insights into special functions that defy conventional understanding.
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Calculus II Lecture #1 Transcendental Functions Civil Engineering Department College of Engineering Mustansiriayah University May 2020 1 Calculus II, Lecture #1 09/05/2025
Transcendental Functions Transcendental Functions The Transcendental functions are those functions which are special, unusual and cannot be understood in ordinary way. One-to-One Functions A function (x) is one-to-one on a domain D if (x1) (x2) whenever x1 x2 in D. Example 1: ? ? = because ?1 ? ? = sin? is not one-to-one on the interval [0,2 ] because sin? 6 ? is one-to-one on any domain of nonnegative numbers ?2whenever x1 x2. 6= sin5? 2 Calculus II, Lecture #1 09/05/2025
Inverse Functions Inverse Functions Definition: Suppose that is a one-to-one function on a domain D with range R. The inverse function -1is defined by: ? 1? = ? if ? ? = ? The domain of -1is the range and the range of -1is the domain. Example 2: Suppose a function y = (x) is given by a table of values: 1 2 3 4 5 6 7 8 x 3 4.5 7 10.5 15 20.5 27 34.5 f(x) 3 Calculus II, Lecture #1 09/05/2025
Inverse Functions Inverse Functions A table for the values of ? = 1(?) can then be obtained by simply interchanging the values in each column of the table for : 3 4.5 7 10.5 15 20.5 27 34.5 y f-1(y) 1 2 3 4 5 6 7 8 Finding Inverses Example 3: Find the inverse of ? =? 2+ 1, expressed as a function of x. Solution: 1. Solve for x in terms of y: ? =? 2+ 1 2? = ? + 2 ? = 2? 2 2. Interchange x and y: ? = 2? 2 4 Calculus II, Lecture #1 09/05/2025
The inverse of the function f(x)=x/2+1 is the function ^(-1) (x) = 2x 2 (see the figure below). To check that, we verify that both composites give the identity function: ? 1? ? ? 2+ 1 2 = ? = 2 =1 ? ? 1? 22? 2 + 1 = ? 5 Calculus II, Lecture #1 09/05/2025
Example 4: Find the inverse of the function ? = ?2, ? 0 expressed as a function of x. Solution: For? 0, the graph satisfies the horizontal line test, so the function is one-to-one and has an inverse. To find the inverse, we first solve for x in terms of y: ? = ?2 ? = ? = ?, because ? 0 We then interchange x and y, obtaining: ? = ? The inverse of the function ? = ?2, is the function ? = ? 6 Calculus II, Lecture #1 09/05/2025
Example 5: Show that ? ?1? = ? 1? ? = ? for: ? ? = ?5 Solution: ? = ?5 ? = ?1/5 ? 1? = ?1/5 Check! = ?1/55= ? ? ? 1? ? 1? ? = ?5 1/5= ?,OK. 7 Calculus II, Lecture #1 09/05/2025
Derivatives of Inverse of Differentiable Functions Derivatives of Inverse of Differentiable Functions 1 ? 1 ? = ? ? 1? ?? 1 ?? 1 or = ?? ???=? 1(?) Example 6: Find the derivative of inverse for ? ? = ?2,? 0 and its value when x=2. Solution: ? ? = ?2,? 0 and its inverse ? 1? = ? ? ? = 2? 1 ? 1 ? = ? ? 1? 1 1 = = 2 ? 1? 2 ? At x=2, ? 2 = 2(2) = 4 and the derivative of ? 1at ?(2), are reciprocals, therefore: 1 ? ? 14 1 2?=1 1 ? 1 4 = = = ? 2 4 8 Calculus II, Lecture #1 09/05/2025
Derivatives of Inverse of Derivatives of Inverse of Differentiable Differentiable Functions Functions Example 7: Find the derivative of inverse for ? ? = 2? + 3, Solution: ? = 2? + 3 ? =? 3 ? 1? =? 3 2 Check! 2 ? 3 2 ? ? 1? = 2 + 3 = ? =2? + 3 3 ? ? = 2 ? 1? ? = ?,OK. 2 1 =1 ? 1 ? = 2 ? ? 1? Check! ? 1 ? =1 2, OK 9 Calculus II, Lecture #1 09/05/2025
Derivatives of Inverse of Differentiable Functions Derivatives of Inverse of Differentiable Functions Example 8: Find the derivative of inverse for ? ? = 2?2,? 0 and its value when x=5. ? 2 Solution: ? ? = 2?2,? 0 and its inverse ? 1? = ? 5 = 2 52= 50 50 ? 150 = = 5 2 ?? 1 ?? 1 ;?? ??= 4? = ?? ???=? 1(?) 1 1 1 = = = 4 5 20 4 ? 1? 10 Calculus II, Lecture #1 09/05/2025
