(6) Find the absolute minimum and maximum of f(x) = x² - 2x¹ on the interval (0,1). (A) There is no absolute minimum or maximum on the interval (B) There is no absolute minimum but fmax = at x = = 1/2 (C) fmin = -1 at x = 1 and fmax = at x = (D) fmin=0 at x = 0 and fmax = 21/12 at x = =1/12 (7) Let f(x)=√4-x2 be a function. For what value of x in the closed interval [-2, 0] does the instantaneous rates of change of f equal the average rate of change of f over that interval? (A) 0 (B) - 1 (C) -√2 (D) -√3 The position function for a particle moving along a line is s(t) = t³-t²-5t + 9, t≥ 0, i meters and t in seconds. For which of the values of t is the particle at rest? (A) t =s and -1s (B) t=1s (C) t = -s and 1 s (D) t = t = 1/3 s Given that f(x) = sin² x - cos x on the closed interval [0, 7], find the absolute maximus minimum value and and state where those values occur.

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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(6) Find the absolute minimum and maximum of f(x) = x² - 2x¹ on the interval (0, 1).
(B) There is no absolute minimum but fmax = at x =
(A) There is no absolute minimum or maximum on the interval
=1/12
(C) fmin=-1 at x = 1 and fmax = at x = 1/2
(D) fmin = 0 at x = 0 and fmax = at x = =1/12
(7) Let f(x) = √4-r2 be a function. For what value of x in the closed interval [-2, 0] does the
instantaneous rates of change of f equal the average rate of change of f over that interval?
(A) 0
(B) - 1
(C) -√2
(D) -√3
(8) The position function for a particle moving along a line is s(t) = t³-t²-5t +9, t≥ 0, in
meters and t in seconds. For which of the values of t is the particle at rest?
(A) t =s and -1s (B) t = 1 s
(C) t = -s and 1 s
0.
5/3
) Given that f(x) = sin² x - cos x on the closed interval [0, π], find the absolute maximus and
minimum value and and state where those values occur.
1 at x = 0
(D) t = s
Transcribed Image Text:(6) Find the absolute minimum and maximum of f(x) = x² - 2x¹ on the interval (0, 1). (B) There is no absolute minimum but fmax = at x = (A) There is no absolute minimum or maximum on the interval =1/12 (C) fmin=-1 at x = 1 and fmax = at x = 1/2 (D) fmin = 0 at x = 0 and fmax = at x = =1/12 (7) Let f(x) = √4-r2 be a function. For what value of x in the closed interval [-2, 0] does the instantaneous rates of change of f equal the average rate of change of f over that interval? (A) 0 (B) - 1 (C) -√2 (D) -√3 (8) The position function for a particle moving along a line is s(t) = t³-t²-5t +9, t≥ 0, in meters and t in seconds. For which of the values of t is the particle at rest? (A) t =s and -1s (B) t = 1 s (C) t = -s and 1 s 0. 5/3 ) Given that f(x) = sin² x - cos x on the closed interval [0, π], find the absolute maximus and minimum value and and state where those values occur. 1 at x = 0 (D) t = s
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