Bradford T. answered • 11/21/22
Tutor
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Retired Engineer / Upper level math instructor
a) f'(x) = 1-2cos(2x)
1-2cos(2x)=0
cos(2x)=1/2
2x = π/3 and 5π/3 on [0,π]
x = π/6, 5π/6
b) f(π/6) = -0.34
f(5π/6) = 3.48
f(0) = 0
f(π) = π
c) absolute max = 3.48
absolute min = -0.34
d) You can graph the function using Desmos or equivalent graphing tool.
Bradford T.
The problem was wanting you to graph f(x)=x-sin(2x) for the interval. If doing manually, the min/max and end points derived will help sketch the function. I think it was mostly asking so that you could visibly see the requested values on the curve. The graph is sort of a distorted sine wave going up a hill.
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11/21/22
Jens Q.
Oh okay, yeah i get it. Since it goes uphill, would the absolute values be diagonal from the interval?
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11/22/22
Jens Q.
What functions would you have to graph? Is it x = pi/6 and 5pi/6, and f(pi) = pi, and the absolute values on the graph?11/21/22