Question

Which of the following is true about the graph of f on the interval [0,pi]? f(x)=x+sin(2x)-2

a) f(x) is increasing on the interval ((1/3)pi, (2/3)pi)
b) f(x) is concave up on the interval (0, (2/3)pi)
c) f(x) is decreasing on the interval (0,pi)
d)f(x) has a local max at the point ((1/3)pi , (1/3)pi+(1/2)sqrt(3)-2)
e)f(x) has a point of inflection at (0,-2)

Answer 1

You need to find first and second derivative of the function and find its behavior in the interval [0,pi]. If f'(x) is positive it is increasing. If f'(x) is negative it is decreasing. If f''(x) > 0 it is concave up. if f"(0) is 0 at (0, -2) then it has a potential point of inflection at that point, etc.

Answer 2

Ok so then the first derivative of f(x) is
\[1+2\cos(2x)\]
and the second derivative is
\[-4\sin(2x)\]

But when I try to solve for zero on the first derivative I dont know exactly which are the critical numbers

\[1+2\cos(2x)=0\]
\[\cos(2x)=\frac{ 1 }{ 2 }\]

Answer 3

cos(2x) = -1/2

Answer 4

I know 1/2=cos(pi/3) but does x just equal pi/6?

Answer 5

or my bad a negetive 1/2

Answer 6

so -1/2 of cosine is 2pi/3 so what is half of that

Answer 7

cos(2pi/3) = -1/2
2x = 2pi/3. x = pi/3.

Answer 8

OH ok that makes much more sense now

Answer 9

So then the critical number of the function is just pi/3

Answer 10

Also cosine repeats itself and so you have to find all solutions in the interval [0, pi] for x.

Answer 11

So then I have to find where in between [0,pi] where it equals zero or its undefined?

Answer 12

I dont know what you mean by how it represents itself

Answer 13

not represents but repeats itself. cosine is a cyclic function and so there will be multiple times it will have a value of -1/2.

Answer 14

Oh ok so their are more critical values than 2pi/3 correct?

Answer 15

or pi/3

Answer 16

Yes.

Answer 17

Ok then so are the cirtical numbers just pi/3 and 2pi/3 or is there more? If so how should I solve for the others?

Answer 18

For cosine there are infinite number of points where it is -1/2. But here the domain is [0, pi] and so x = pi/3 and 2pi/3 are the two (potential) critical points.
At critical points, the function has a max/min. Evaluate second derivative at those two points.

Answer 19

Ok then

\[-4\sin(2(\frac{ \pi }{ 3 })<0\]

and

\[-4\sin(2(\frac{ 2\pi }{ 3 })>0\]

Answer 20

So pi/3 is a max and 2pi/3 is a min

Answer 21

Yes, pi/3 is a max and 2pi/3 is a min. Therefore, we can infer the function is increasing in the interval 0, pi/3 (in order to reach max at pi/3) and decreasing in the interval pi/3, 2pi/3 (to reach min at 2pi/3).

Answer 22

So my awnser choices are
a) f(x) is increasing on the interval ((1/3)pi, (2/3)pi)
b) f(x) is concave up on the interval (0, (2/3)pi)
c) f(x) is decreasing on the interval (0,pi)
d)f(x) has a local max at the point ((1/3)pi , (1/3)pi+(1/2)sqrt(3)-2)
e)f(x) has a point of inflection at (0,-2)

So (a) is not correct because pi/3 is a max and 2pi/3 is a min so it has to decrease between those two intervals.
B is not correct because pi/3 is concave down
C is not correct because their are points where f(x) increases
D may be correct because pi/3 is a max but I dont know how to get to that y value
E im not sure of either

Answer 23

So far Im guessing its D

Answer 24

Because I chose E on the quiz and it was wrong

Answer 25

Your conclusions about a, b and c above are correct.
For d, you know the max is at x = pi/3. Put x = pi/3 and evaluate f(x).

Answer 26

So
\[f(\frac{ \pi }{ 3 })=\frac{ \pi }{ 3 }+\sin(2(\frac{ \pi }{ 3 })-2\]

Answer 27

that equals
\[\frac{ \pi }{ 3 }+\frac{ \sqrt(3) }{ 2 }-2\]

Answer 28

Oh ok its simpler than Im thinking it is

Answer 29

Thank you so much ranga and thanks for being patient

Answer 30

You are welcome Matthew. You did most of it yourself. Kudos.