Critical Points/Concavity of Trig Functions

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pepperonibread

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Hi everyone, in my class we've been finding increasing/decreasing intervals, concavity, and inflection points, and two trig problems have me especially confused:

1. f(x)=sinx-cosx, on the interval [-pi,pi]. For this one I'm looking for inflection points, so i took the 2nd derivative: f''(x)=-sinx+cosx. From there, I set it equal to zero, and solved like so:
-sinx+cosx=0
(cosx-sinx)[sup:1f1hoik9]2[/sup:1f1hoik9]=0[sup:1f1hoik9]2[/sup:1f1hoik9]
cos[sup:1f1hoik9]2[/sup:1f1hoik9]x-sin[sup:1f1hoik9]2[/sup:1f1hoik9]x=0
cos[sup:1f1hoik9]2[/sup:1f1hoik9]x+sin[sup:1f1hoik9]2[/sup:1f1hoik9]x=2sin[sup:1f1hoik9]2[/sup:1f1hoik9]x (added 2sin[sup:1f1hoik9]2[/sup:1f1hoik9]x to both sides)
1=2sin[sup:1f1hoik9]2[/sup:1f1hoik9]x
1/2=sin[sup:1f1hoik9]2[/sup:1f1hoik9]x
sqrt(1/2)=sinx
(sqrt(2))/2=sinx
x=pi/4 and 3pi/4
actually x=plus or minus pi/4 and 3pi/4 because I took the square root.
Now the problem comes in when i look at the actually graph of the function... from the graph it's easy to see that there actually only inflection points at negative 3pi/4 and positive pi/4. So what's the issue with my math here? I'm not sure why I'm getting 4 answers when I only need 2.

2. f(x)=(sin(2x))[sup:1f1hoik9]2[/sup:1f1hoik9], on the interval [0, pi]. Here I'm looking for intervals of increasing/decreasing, so I took the 1st derivative by the chain rule: f'(x)=2sin(2x)*cos(2x)*2, or f'(x)=4sin(2x)cos(2x). I set the derivative equal to zero to find critical points, and solved like so:
4sin(2x)cos(2x)=0
sin(2x)cos(2x)=0
2sinxcosx(cos[sup:1f1hoik9]2[/sup:1f1hoik9]x-sin[sup:1f1hoik9]2[/sup:1f1hoik9]x)=0 (double angle formulas)
sinxcosx(cos[sup:1f1hoik9]2[/sup:1f1hoik9]x-sin[sup:1f1hoik9]2[/sup:1f1hoik9]x)=0
sinxcos[sup:1f1hoik9]3[/sup:1f1hoik9]x-sin[sup:1f1hoik9]3[/sup:1f1hoik9]xcosx=0
sinxcos[sup:1f1hoik9]3[/sup:1f1hoik9]x+sin[sup:1f1hoik9]3[/sup:1f1hoik9]xcosx=2sin[sup:1f1hoik9]3[/sup:1f1hoik9]xcosx (added 2sin[sup:1f1hoik9]3[/sup:1f1hoik9]xcosx to both sides)
sinxcosx(cos[sup:1f1hoik9]2[/sup:1f1hoik9]x+sin[sup:1f1hoik9]2[/sup:1f1hoik9]x)=2sin[sup:1f1hoik9]3[/sup:1f1hoik9]xcosx
sinxcosx(1)=2sin[sup:1f1hoik9]3[/sup:1f1hoik9]xcosx
1=2sin[sup:1f1hoik9]2[/sup:1f1hoik9]x
1/2=sin[sup:1f1hoik9]2[/sup:1f1hoik9]x
sqrt(1/2)=sinx
(sqrt(2))/2=sinx
x=pi/4 and 3pi/4 (not plus or minus in this case because the interval is [0,pi]).
However, on this one I have the opposite problem as before. Looking at the graph of the original function i can see that there are actually five critical points: 0, pi/4, pi/2, 3pi/4, and pi. I feel like the problem with this may have to do with using the double angle identities, though I'm still not sure how to get all five solutions.

Anyway, if anyone could point me in the right direction on these it would be greatly appreciated - thanks.
 
pepperonibread said:
Hi everyone, in my class we've been finding increasing/decreasing intervals, concavity, and inflection points, and two trig problems have me especially confused:

1. f(x)=sinx-cosx, on the interval [-pi,pi]. For this one I'm looking for inflection points, so i took the 2nd derivative: f''(x)=-sinx+cosx. From there, I set it equal to zero, and solved like so:
-sinx+cosx=0
(cosx-sinx)[sup:ufsxn6tz]2[/sup:ufsxn6tz]=0[sup:ufsxn6tz]2[/sup:ufsxn6tz]
cos[sup:ufsxn6tz]2[/sup:ufsxn6tz]x-sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x=0 <<<<<<<<<<<<<<<<<That is wrong

-sinx+cosx=0

sin(x) = cos(x)

sin(x) = sin(?/2 - x) or sin(x) = sin(-3?/2 - x)

x = ?/4 or -3?/4

cos[sup:ufsxn6tz]2[/sup:ufsxn6tz]x+sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x=2sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x (added 2sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x to both sides)
1=2sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x
1/2=sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x
sqrt(1/2)=sinx
(sqrt(2))/2=sinx
x=pi/4 and 3pi/4
actually x=plus or minus pi/4 and 3pi/4 because I took the square root.
Now the problem comes in when i look at the actually graph of the function... from the graph it's easy to see that there actually only inflection points at negative 3pi/4 and positive pi/4. So what's the issue with my math here? I'm not sure why I'm getting 4 answers when I only need 2.

2. f(x)=(sin(2x))[sup:ufsxn6tz]2[/sup:ufsxn6tz], on the interval [0, pi]. Here I'm looking for intervals of increasing/decreasing, so I took the 1st derivative by the chain rule: f'(x)=2sin(2x)*cos(2x)*2, or f'(x)=4sin(2x)cos(2x). I set the derivative equal to zero to find critical points, and solved like so:
4sin(2x)cos(2x)=0
sin(2x)cos(2x)=0
2sinxcosx(cos[sup:ufsxn6tz]2[/sup:ufsxn6tz]x-sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x)=0 (double angle formulas)
sinxcosx(cos[sup:ufsxn6tz]2[/sup:ufsxn6tz]x-sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x)=0
sinxcos[sup:ufsxn6tz]3[/sup:ufsxn6tz]x-sin[sup:ufsxn6tz]3[/sup:ufsxn6tz]xcosx=0
sinxcos[sup:ufsxn6tz]3[/sup:ufsxn6tz]x+sin[sup:ufsxn6tz]3[/sup:ufsxn6tz]xcosx=2sin[sup:ufsxn6tz]3[/sup:ufsxn6tz]xcosx (added 2sin[sup:ufsxn6tz]3[/sup:ufsxn6tz]xcosx to both sides)
sinxcosx(cos[sup:ufsxn6tz]2[/sup:ufsxn6tz]x+sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x)=2sin[sup:ufsxn6tz]3[/sup:ufsxn6tz]xcosx
sinxcosx(1)=2sin[sup:ufsxn6tz]3[/sup:ufsxn6tz]xcosx
1=2sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x
1/2=sin[sup:ufsxn6tz]2[/sup:ufsxn6tz]x
sqrt(1/2)=sinx
(sqrt(2))/2=sinx
x=pi/4 and 3pi/4 (not plus or minus in this case because the interval is [0,pi]).
However, on this one I have the opposite problem as before. Looking at the graph of the original function i can see that there are actually five critical points: 0, pi/4, pi/2, 3pi/4, and pi. I feel like the problem with this may have to do with using the double angle identities, though I'm still not sure how to get all five solutions.

Anyway, if anyone could point me in the right direction on these it would be greatly appreciated - thanks.
Click to expand...
 
Subhotosh Khan said:
1. f(x)=sinx-cosx, on the interval [-pi,pi]. For this one I'm looking for inflection points, so i took the 2nd derivative: f''(x)=-sinx+cosx. From there, I set it equal to zero, and solved like so:
-sinx+cosx=0
(cosx-sinx)[sup:2qlaljsc]2[/sup:2qlaljsc]=0[sup:2qlaljsc]2[/sup:2qlaljsc]
cos[sup:2qlaljsc]2[/sup:2qlaljsc]x-sin[sup:2qlaljsc]2[/sup:2qlaljsc]x=0 <<<<<<<<<<<<<<<<<That is wrong

-sinx+cosx=0

sin(x) = cos(x)

sin(x) = sin(?/2 - x) or sin(x) = sin(-3?/2 - x)

x = ?/4 or -3?/4
Click to expand...

oh duh of course (cosx-sinx)[sup:2qlaljsc]2[/sup:2qlaljsc] != cos[sup:2qlaljsc]2[/sup:2qlaljsc]x-sin[sup:2qlaljsc]2[/sup:2qlaljsc]x... basic foil mistake :roll: . thanks!
would you happen to have any insight into the second problem?
 
2. f(x)=(sin(2x))[sup:3r967e07]2[/sup:3r967e07], on the interval [0, pi]. Here I'm looking for intervals of increasing/decreasing, so I took the 1st derivative by the chain rule: f'(x)=2sin(2x)*cos(2x)*2, or f'(x)=4sin(2x)cos(2x). I set the derivative equal to zero to find critical points, and solved like so:
4sin(2x)cos(2x)=0
sin(2x)cos(2x)=0
2sinxcosx(cos[sup:3r967e07]2[/sup:3r967e07]x-sin[sup:3r967e07]2[/sup:3r967e07]x)=0 (double angle formulas)
sinxcosx(cos[sup:3r967e07]2[/sup:3r967e07]x-sin[sup:3r967e07]2[/sup:3r967e07]x)=0
sinxcos[sup:3r967e07]3[/sup:3r967e07]x-sin[sup:3r967e07]3[/sup:3r967e07]xcosx=0
sinxcos[sup:3r967e07]3[/sup:3r967e07]x+sin[sup:3r967e07]3[/sup:3r967e07]xcosx=2sin[sup:3r967e07]3[/sup:3r967e07]xcosx (added 2sin[sup:3r967e07]3[/sup:3r967e07]xcosx to both sides)
sinxcosx(cos[sup:3r967e07]2[/sup:3r967e07]x+sin[sup:3r967e07]2[/sup:3r967e07]x)=2sin[sup:3r967e07]3[/sup:3r967e07]xcosx
sinxcosx(1)=2sin[sup:3r967e07]3[/sup:3r967e07]xcosx
1=2sin[sup:3r967e07]2[/sup:3r967e07]x
1/2=sin[sup:3r967e07]2[/sup:3r967e07]x
sqrt(1/2)=sinx
(sqrt(2))/2=sinx
x=pi/4 and 3pi/4 (not plus or minus in this case because the interval is [0,pi]).
However, on this one I have the opposite problem as before. Looking at the graph of the original function i can see that there are actually five critical points: 0, pi/4, pi/2, 3pi/4, and pi. I feel like the problem with this may have to do with using the double angle identities, though I'm still not sure how to get all five solutions.

Anyway, if anyone could point me in the right direction on these it would be greatly appreciated - thanks.
Click to expand...

You can easily see the intervals of increase and decrease on the graph:

But, \(\displaystyle f(x)=sin^{2}(2x)\)

\(\displaystyle f'(x)=4sin(2x)cos(2x)\)

Setting this to 0 and solving for x gives:

\(\displaystyle x=\frac{{\pi}C}{2}-\frac{\pi}{4}, \;\ x=\frac{C{\pi}}{2}\)

The maximums are on the left and the minimums on the right .

So, in between is where it increases and decreases.

Check the derivatives. If they are negative, then it is decreasing. If they are positive, then increasing.

Increasing from 0 to \(\displaystyle \frac{\pi}{4}\)

Decreasing from \(\displaystyle \frac{\pi}{4} \;\ to \;\ \frac{\pi}{2}\)

Increasing from \(\displaystyle \frac{\pi}{2} \;\ to \;\ \frac{3\pi}{4}\)

Decreasing from \(\displaystyle \frac{3\pi}{4} \;\ to \;\ {\pi}\)

And so on.
 
Hello, pepperonibread!

\(\displaystyle 1.\;f(x)\:=\:\sin x-\cos x,\:\text{ on the interval }\,[-\pi,\:\pi].\)
. . . \(\displaystyle \text{Find the inflection points.}\)

\(\displaystyle \text{i took the 2nd derivative: }\,f"(x)\:=\:-\sin x + \cos x\)
. . . \(\displaystyle \text{ and set it equal to zero: }\;-\sin x + \cos x\:=\:0\)
Click to expand...

\(\displaystyle \text{We have: }\;\sin x \:=\:\cos x \;\;\;\Rightarrow\;\;\;\frac{\sin x }{\cos x} \:=\:1 \;\;\;\Rightarrow\;\;\;\tan x \:=\:1\)

\(\displaystyle \text{Therefore: }\;x \:=\:\frac{\pi}{4},\:-\frac{3\pi}{4}\)

\(\displaystyle \text{The inflection points are: }\;\left(\frac{\pi}{4},\,0\right)\:\text{ and }\:\left(-\frac{3\pi}{4},\,0\right)\)

 
Hello again, pepperonibread!

\(\displaystyle 2.\;f(x)\:=\:\sin^22x\:\text{ on the interval }[0, \pi]\)
. . . \(\displaystyle \text{Find intervals of increasing/decreasing.}\)

\(\displaystyle \text{I took the 1st derivative by the chain rule: }\:f'(x)\:=\:2\sin 2x\cos2x\cdot 2 \:=\:4\sin2x\cos2x\)
\(\displaystyle \text{I set the derivative equal to zero to find critical points . . .}\)
Click to expand...

\(\displaystyle \text{We have: }\:4\sin2x\cos2x \:=\:0 \;\;\;\Rightarrow\;\;\;2\underbrace{(2\sin2x\cos2x)}_{\text{This is }\sin4x} \:=\:0 \;\;\;\Rightarrow\;\;\;2\sin4x \:=\:0\)

\(\displaystyle \text{So we have: }\:\sin4x \:=\:0 \;\;\;\Rightarrow\;\;\;4x \:=\:0,\pi,\:2\pi,\:3\pi,\:4\pi,\:5\pi,\:\hdots\)

. . \(\displaystyle \text{Hence: }\:x \:=\:0,\:\frac{\pi}{4},\:\frac{\pi}{2},\:\frac{3\pi}{4},\:\pi\)


\(\displaystyle \text{Therefore: }\;\begin{Bmatrix} \text{Increasing} & \left(0,\:\dfrac{\pi}{4}\right) \cup \left(\dfrac{\pi}{2},\:\dfrac{3\pi}{4}\right) \\ \\[-2mm]\text{Decreasing} & \left(\dfrac{\pi}{4},\:\dfrac{\pi}{2}\right) \cup \left(\dfrac{3\pi}{4},\:\pi\right) \end{Bmatrix}\)
 
\(\displaystyle 1) \ f(x) \ = \ sin(x)-cos(x), \ find \ inflection \ points \ for \ interval \ [-\pi,\pi].\)

\(\displaystyle Hence, \ f"(x) \ = \ -sin(x)+cos(x), \ set \ to \ zero \ gives: \ sin(x) \ = \ cos(x).\)

\(\displaystyle This \ implies \tan(x) \ = \ 1 \ [divide \ both \ sides \ by \cos(x)], \ x \ = \ \frac{\pi}{4} \ + \ k\pi, \ k \ an \ integer.\)

\(\displaystyle Now, \ since \ our \ interval \ is \ restricted \ to \ [-\pi,\pi], \ x \ = \ \frac{\pi}{4}, \ \frac{-3\pi}{4}.\)

\(\displaystyle Therefore, \ our \ two \ points \ of \ inflection \ are \ f(\pi/4) \ = \ 0 \ and \ f(-3\pi/4) \ = \ 0, \ see \ graph.\)

[attachment=0:2s3ndcwu]ggg.jpg[/attachment:2s3ndcwu]
 
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