Explanation Given: The given function is y = x 3 − 24 ln x . Calculation: Given: the given interval is [ 1 2 , 3 ] . Use sum/ difference rule : ( f − g ) ' = f ′ − g ′ y = x 3 − 24 ln x y ' = d d x ( x 3 ) − 24 d d x ( ln x ) y ' = 3 x 2 − 24 x For critical point y ' = 0 . 3 x 2 − 24 x = 0 ⇒ 3 x 3 − 24 x = 0 ⇒ 3 x 3 − 24 = 0 ⇒ 3 x 3 − 24 + 24 = 0 + 24 ⇒ 3 x <

4th Edition

ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Question

Chapter 4.2, Problem 60E

To determine

The minimum and maximum values of the given function in the given interval by comparing with critical and end points.

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Chapter 4.2, Problem 59E

Chapter 4.2, Problem 61E

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In Exercises 13-14, find the domain of each function. 13. f(x) 3 (х +2)(х — 2) 14. g(x) (х + 2)(х — 2) In Exercises 15–22, let f(x) = x? – 3x + 8 and g(x) = -2x – 5.

Please explain as detail as possible, thanks.

Each of Exercises 25–36 gives a formula for a function y = f(x). In each case, find f-x) and identify the domain and range of f-. As a check, show that f(fx)) = f-"f(x)) = x. 25. f(x) = x 26. f(x) = x, x20 %3D %3D 27. f(x) = x + 1 28. f(x) = (1/2)x – 7/2 30. f(x) = 1/r, x * 0 %3D 29. f(x) = 1/x, x>0 x + 3 31. f(x) 32. f(x) = VE - 3 34. f(x) = (2x + 1)/5 2 33. f(x) = x - 2r, xs1 (Hint: Complete the square.) * + b x - 2' 35. f(x) = b>-2 and constant 36. f(x) = x? 2bx, b> 0 and constant, xsb

Chapter 4 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

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The minimum and maximum values of the given function in the given interval by comparing with critical and end points.

Solution Summary: The author compares the minimum and maximum values of a given function in the given interval by comparing with critical and end points.

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The minimum and maximum values of the given function in the given interval by comparing with critical and end points.

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Chapter 4 Solutions