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(i) f(x) = x³ – x² - 8x + 1 on [-2, 2]. (ii) f(x) = x³ + x + 1 (iii) f(x) on [-1, 1]. = 3x* – 8x³ + 6x² on [-}, }].

(i) f(x) = x³ – x² - 8x + 1 on [-2, 2]. (ii) f(x) = x³ + x + 1 (iii) f(x) on [-1, 1]. = 3x* – 8x³ + 6x² on [-}, }].

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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8:37 AM Ø VOLTE E 4 28% (Once again, the reader is invited to supply the details of this part of the argument.) I PROBLEMS 1. For each of the following functions, find the maximum and minimum values on the indicated intervals, by finding the points in the interval where the derivative is 0, and comparing the values at these points with the values at the end points. (i) f(x) = x³ – x² - 8x + 1 on [-2, 2]. (ii) f(x) = x³ + x + 1 (iii) f(x) = 3x* – 8x³ + 6x² on [-1, 1]. on (-}, }). %3D 1 (iv) f(x) on [-, 1]. xi + x + 1 x + 1 (v) f(x) on [-1, }). = x? + 1 (vi) f(x) on [0, 5]. х? — 1 11. Significance of the Derivative 191 2. Now sketch the graph of each of the functions in Problem 1, and find all local maximum and minimum points. 3. Sketch the graphs of the following functions. (i), f(x) = x + 1 3 (ii) ƒ(x) = x + x² x2 (iii) f(x) x? – 1 1 (iv) f(x) 1 + x? 4. (a) If a1 < · · · < an, find the minimum value of f(x) = } (x – a;)². i=1 *(b) Now find the minimum value of f(x) = ) x – a;|. This is a prob- lem where calculus won't help at all: on the intervals between the a's the function f is linear, so that the minimum clearly occurs at one of the ai, and these are precisely the points where f is not differ- entiable. However, the answer is easy to find if you consider how f(x) changes as you pass from one such interval to another. *(c) Let a > 0. Show that the maximum value of 1 1 f(x) 1 + |x| 1 + |x - a is (2 + a)/(1 + a). (The derivative can be found on each of the intervals (- o, 0), (0, a), and (a, ∞) separately.)
Transcribed Image Text:8:37 AM Ø VOLTE E 4 28% (Once again, the reader is invited to supply the details of this part of the argument.) I PROBLEMS 1. For each of the following functions, find the maximum and minimum values on the indicated intervals, by finding the points in the interval where the derivative is 0, and comparing the values at these points with the values at the end points. (i) f(x) = x³ – x² - 8x + 1 on [-2, 2]. (ii) f(x) = x³ + x + 1 (iii) f(x) = 3x* – 8x³ + 6x² on [-1, 1]. on (-}, }). %3D 1 (iv) f(x) on [-, 1]. xi + x + 1 x + 1 (v) f(x) on [-1, }). = x? + 1 (vi) f(x) on [0, 5]. х? — 1 11. Significance of the Derivative 191 2. Now sketch the graph of each of the functions in Problem 1, and find all local maximum and minimum points. 3. Sketch the graphs of the following functions. (i), f(x) = x + 1 3 (ii) ƒ(x) = x + x² x2 (iii) f(x) x? – 1 1 (iv) f(x) 1 + x? 4. (a) If a1 < · · · < an, find the minimum value of f(x) = } (x – a;)². i=1 *(b) Now find the minimum value of f(x) = ) x – a;|. This is a prob- lem where calculus won't help at all: on the intervals between the a's the function f is linear, so that the minimum clearly occurs at one of the ai, and these are precisely the points where f is not differ- entiable. However, the answer is easy to find if you consider how f(x) changes as you pass from one such interval to another. *(c) Let a > 0. Show that the maximum value of 1 1 f(x) 1 + |x| 1 + |x - a is (2 + a)/(1 + a). (The derivative can be found on each of the intervals (- o, 0), (0, a), and (a, ∞) separately.)
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