1. (a) Use interval notation to indicate where (in that interval) the graph of g
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Question
1.
(a) Use interval notation to indicate where (in that interval) the graph of g is concave up: ?
(b) Use interval notation to indicate where (in that interval) the graph of g is concave down:?
(c) The inflection points of g, in the interval [0,pi], are: ?
2.
(If there is more than one inflection point, enter them as a comma separated list.)
3.
(a) what are the critical point(s)? :
(b) what are the x-coordinates of the inflection Point(s)? :
1 Analyze function g(theta) = 5 theta + 4 (sin(theta))^2 on the interval [ 0 , pi ]. (a) Use interval notation to indicate where (in that interval) the graph of g is concave up: ? (b) Use interval notation to indicate where (in that interval) the graph of g is concave down:? (c) The inflection points of g, in the interval [0,pi], are: ? 2 The figure below is the graph of the derivative f' of a function f. BLE245-433-setC4-Q-4-Monot-Concprob6imag (x values are 0 when x=0, 8 and 17 on the graph) (a) Give the x-coordinate(s) of the inflection point(s) of f. : ? (If there is more than one inflection point, enter them as a comma separated list.) (b) On which interval(s) is fconcave down? 3 f(x) = 5x+5sin!left(xright), 0 le x le 2 pi (a) what are the critical point(s)? : (b) what are the x-coordinates of the inflection Point(s)? : 4. A particle moves along a straight line and its position at time t is given by s(t)= 2t^3 - 27 t^2 + 108 t where s is measured in feet and t in seconds. (a) what is the TOTAL distance the particle travels between time 0 and time 18? :Explanation / Answer
1.
i am taking theta as t
g(t) = 5t + 4sin2t
g'(t) = 5 + 8sintcost
g'(t) = 5 + 4sin2t
g''(t) = 8cos2t
(a)
for concave up, g''(t) > 0
8cos2t > 0
cos2t > 0
0 < 2t < pi/2 and 3pi/2 < t < 2pi
0 < t < pi/4 and 3pi/4 < t < pi..........concave up
(b)
for concave down, interval will be
8cos2t < 0
cos2t < 0
pi/2 < 2t < 3pi/2 and
?pi/4 < t < 3pi/4---------concave down
(c)
for inflection points, g''(t) = 0
8cos2t = 0
cos2t = 0
2t = pi/2, 3pi/2
t = pi/4, 3pi/3..........inflection point
2.
(a)
from the graph we have horizontal tangents at x = 6 and x = 13
Hence inflection points are at x = 6, 13
(b)
f(x) is concave down from 6 < x < 13
3.
f(x) = 5x + 5sinx
f'(x) = 5 + 5cosx
f''(x) = -5sinx
(a)
for critical points, f'(x) = 0
5 + 5cosx = 0
cosx = -1
x = pi
(b)
for inflection points, f''(x) = 0
-5sinx = 0
x = 0, pi, 2pi
4.
s(t) = 2t3 - 27t2 + 108t
total distance travelled between t (0) - t(18) will be
S = s(18) - s(0)
S = 2*183 - 27*182 + 108*18
S = 11664 - 8748 + 1944
S = 4860 feet
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