Finding points on a trigonometric function

[meeklobraca]

Homework Statement

Use the function f(x) = sin x + cos x, where 0<=x<=pi to answer the following.

a) Determine all stationary points. Classify the points as minimum or maximum
b) locate the points of inflection
c) determine the endpoints of the interval

Homework Equations

The Attempt at a Solution

So I know that the first and second derivative are

f1(x) = cos x - sin x
f2(x) = -sin x - cos x

Im lost at how to use this to determine my values.

Thank you for your help!

 

[Dick]

You are trying to solve e.g. cos(x)-sin(x)=0. Dividing both sides by cos(x) gives 1-tan(x)=0. Is that any help?

 

[meeklobraca]

I just found where cos x = sin x, put that into the function and found my value. I've got the 1) and the 3) of the question. Now I am just stuck on the point of inflection.

Im curious though, can i divide both sides by cos x for the second derivative to find my points of inflection?

If so that would leave me with tan x. How do I determine the point of inflection without actually pluggin in every value from 0 to pi?

 

[Dick]

Sure. Then you get tan(x)=-1 instead of +1, right?

 

[meeklobraca]

Yup your right, so to find the point of inflection should I just do, tan x = -1 and find out what x is?

 

[Dick]

I think so, don't you?

 

[meeklobraca]

Kind of, I get x = -0.8. But isn't that out of range of my domain since it has to be between 0 and pi?

 

[Dick]

I think I can find a value of x between 0 and pi where tan(x)=(-1). Can't you?

 

[meeklobraca]

Yeah I got 3pi/4. Plug that into my function and I got a y = 1?

So my point of inflection is (3pi/4, 1)?

 

[Dick]

I don't think sin(3pi/4)+cos(3pi/4)=1.

 

[meeklobraca]

im sorry, i was in degree mode for this. In radian mode I get -0.1. Correct?

 

[Dick]

im sorry, i was in degree mode for this. In radian mode I get -0.1. Correct?

No again. You don't NEED a calculator. To get f'' you solved sin(x)=(-cos(x)). That should make it pretty easy to figure out f(x)=sin(x)+cos(x).

 

[meeklobraca]

Well no apprantly its not that pretty easy to figure out. Your not making much sense there.

 

[Dick]

I'm trying to point out that in this problem f''(x)=(-f(x)).

 

[lurflurf]

helps to simplify first

f(x) = sin(x)+cos(x)=sqrt(2)sin(x+pi/4)

 

[meeklobraca]

I don't get what you guys are saying here. Whats the problem with plugging in the values I find and using the calculator? "f''(x)=(-f(x))." is no different than me pluggin in
3pi/4 into the original function.

Either way I don't get how I can avoid using a calculator. Even using f''(x)=(-f(x)), f"(x)=0.1, so -f(x) = -0.1 So I get my point of inflection at (3pi/4,-0.1)

 

[HallsofIvy]

The difficulty appears to be that you have no idea what "sine" and "cosine" are and Dick and lurflurf have been assuming that you did.

If you did, then you would know that $sin(3\pi/4)= -\sqrt{2}/2$ and $cos(3\pi/4)= -\sqrt{2}{2}$ so that $f(3\pi/4)= sin(3\pi/4)+ cos(\3pi/4)= -\sqrt{2}$, not "-0.1".

 

[Dick]

sin(3pi/4)=sqrt(2)/2, cos(3pi/4)=(-sqrt(2))/2. Even if you insist on the calculator bit, your's might need a tuneup. I get sin(3pi/4)=0.70710678118655.
cos(3pi/4)=(-0.70710678118655). NOW do you see why the sum isn't '-.1'?

 

[meeklobraca]

Yup, your right I do see that now. The answer is 0. I don't get why my calculator is spitting out -0.1 but I get it now. Thank you!