Let f(x) (a) The derivative of ƒ(x) is ƒ'(x) = 6x^2+6x-336 = 2x³ + 3x² - 336x. Complete this problem without a graphing calculator. (b) As a comma-separated list, the critical points of fare x = Since f is continuous on the closed interval [–9, 15], ƒ has both an absolute maximum and an absolute minimum on the interval [-9, 15] according to the Extreme Value Theorem. To find the extreme values, we evaluate f at the endpoints and at the critical points. -8,7 (c) As a comma-separated list, the y-values corresponding to the critical points and endpoints are y = -8,7 (d) The minimum value of for I 0 15lic 1510
Let f(x) (a) The derivative of ƒ(x) is ƒ'(x) = 6x^2+6x-336 = 2x³ + 3x² - 336x. Complete this problem without a graphing calculator. (b) As a comma-separated list, the critical points of fare x = Since f is continuous on the closed interval [–9, 15], ƒ has both an absolute maximum and an absolute minimum on the interval [-9, 15] according to the Extreme Value Theorem. To find the extreme values, we evaluate f at the endpoints and at the critical points. -8,7 (c) As a comma-separated list, the y-values corresponding to the critical points and endpoints are y = -8,7 (d) The minimum value of for I 0 15lic 1510
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Let f(x) = 2x³ + 3x² - 336x. Complete this problem without a graphing calculator.
(a) The derivative of f(x) is f'(x)
= 6x^2+6x-336
(b) As a comma-separated list, the critical points of f are x =
Since f is continuous on the closed interval [−9, 15], ƒ has both an absolute maximum and an absolute minimum on the interval [-9, 15]
according to the Extreme Value Theorem. To find the extreme values, we evaluate f at the endpoints and at the critical points.
(c) As a comma-separated list, the y-values corresponding to the critical points and endpoints are y = -8,7
(d) The minimum value of f on [-9, 15] is y
the minimum value occurs at x = 7
and this x is a(n) critical point
V
-8,7
= -1519
(e) The maximum value of ƒ on [-9, 15] is y = 1856
the maximum value occurs at x = -8
and this x is a(n) end point](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F418fd126-7f9b-4db4-a312-2e5f96462821%2Fc44819a7-435a-416b-ba2d-4e527ca061e6%2Fcjow8xi_processed.png&w=3840&q=75)
Transcribed Image Text:Let f(x) = 2x³ + 3x² - 336x. Complete this problem without a graphing calculator.
(a) The derivative of f(x) is f'(x)
= 6x^2+6x-336
(b) As a comma-separated list, the critical points of f are x =
Since f is continuous on the closed interval [−9, 15], ƒ has both an absolute maximum and an absolute minimum on the interval [-9, 15]
according to the Extreme Value Theorem. To find the extreme values, we evaluate f at the endpoints and at the critical points.
(c) As a comma-separated list, the y-values corresponding to the critical points and endpoints are y = -8,7
(d) The minimum value of f on [-9, 15] is y
the minimum value occurs at x = 7
and this x is a(n) critical point
V
-8,7
= -1519
(e) The maximum value of ƒ on [-9, 15] is y = 1856
the maximum value occurs at x = -8
and this x is a(n) end point
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