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F(x)=0, 2 - (cscx)^2 = 0

  • #1

Homework Statement



Analyze the function f(x) = 2x + cotx over the interval [0,pi] and state the increases/decreases and concavity.

Homework Equations



I already found the first and second derivatives. However, I'm having trouble when setting them equal to zero.

2 - csc2x = 0

and

2(cscx)^2(cotx)^2 = 0

The Attempt at a Solution



2 - cscx^2 = 0

cscx^2 = 2

maybe 1/sinx^2 = 2

maybe (1/sinx)^2 = 2

and then i'm stuck...
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
Pause with 1/(sin(x)^2)=2. Use algebra to rearrange that to sin(x)^2=1/2. Do you know how to do that? I'm not sure why you are getting stuck. You are doing fine so far.
 
  • #3
Thanks for the help. Yea i was doubting whether I should isolate sinx or just x. I was able to arrange the equation and solve for the angles to get x = pi/4, 3pi/4 since the interval is [0,pi].

However, I'm getting stuck when trying to find the inflection points. The second derivative is 2cosx/sin(x)^3. I know cosx/sinx is cotx. But the power of 3 in the denominator is throwing me off. Also, I know is easier to find the angles of sinx and cosx than cotx, but is it necessary for the equation to be isolated into one trig term, such as cotx = (stuff), or can this be arranged to where sinx and/or cosx = (stuff)?

Ultimately, I am stuck at f"(x) = 2cosx/sin(x)^3 = 0 with little to no idea of what to do.
 
  • #4
1,291
0
Thanks for the help. Yea i was doubting whether I should isolate sinx or just x. I was able to arrange the equation and solve for the angles to get x = pi/4, 3pi/4 since the interval is [0,pi].

However, I'm getting stuck when trying to find the inflection points. The second derivative is 2cosx/sin(x)^3. I know cosx/sinx is cotx. But the power of 3 in the denominator is throwing me off. Also, I know is easier to find the angles of sinx and cosx than cotx, but is it necessary for the equation to be isolated into one trig term, such as cotx = (stuff), or can this be arranged to where sinx and/or cosx = (stuff)?

Ultimately, I am stuck at f"(x) = 2cosx/sin(x)^3 = 0 with little to no idea of what to do.
[tex]\frac{2\cos x}{\sin^3 x}[/tex]

You are trying to differentiate this right?

There are a couple ways you can do this. Start off with the constant rule.

[tex] 2 (\frac{\cos x}{\sin x^3}[/tex]

Hint: [itex]\frac{a}{b^2} = \frac{a}{b}\frac{1}{b} or \frac{1}{x} = x^{-1}[/itex]

There is also a simple differentiation rule that is very helpful in calculus I. Rings a bell?
 
  • #5
[tex]\frac{2\cos x}{\sin^3 x}[/tex]

You are trying to differentiate this right?
No sorry, that is already the second derivative of the original function I posted.

I am trying to solve f"(x) = 0 for f"(x) = 2cosx/sin(x)^3 = 0
 
  • #6
eumyang
Homework Helper
1,347
10
Take the denominator and set it equal to 0. The values of x you find will be values that CANNOT be in the solution set.

Then take the numerator and set it equal to 0. Remove from this solution set any values of x that also make the denominator equal to 0, if there are any.
 
  • #7
So 2cosx = 0, so x should be 2(∏/2) and 2(3∏/2).

so x = ∏ and 3∏

Since the interval given in the problem is [0,∏],

the only allowable x value is x = ∏

-------------------------------------

Now setting (sinx)^3 = 0

I would guess sinx = 0, x = 1 and -1 and since its to the 3rd power,

the x values are x = (1^3 and -1^3) or maybe ([itex]\sqrt[3]{1}[/itex] and[itex]\sqrt[3]{-1}[/itex])

Are either of these correct?

So since none of the denominator roots = the numerator roots, then it stands that the only x value to make 2cosx/(sinx)^3 = 0 within interval [0,∏] is:

x = ∏

And if this is the correct answer, is the way I arrived at that answer correct?

And if its not correct, how do I solve and think about this problem?
 
  • #8
Wait a minute...

2 should not be included in setting cosx = 0

because 2 can be made a seperate term and set equal to zero, such as 2≠0

So I just have to set cosx = 0, which is x = ∏/2 and 3∏/2

And since they still don't match the denominator's roots,

x = ∏/2 is my revised guess.

If this is correct, is my method/reasoning correct?

If not, why and how do I get the correct answer?
 
  • #9
eumyang
Homework Helper
1,347
10
Now setting (sinx)^3 = 0

I would guess sinx = 0, x = 1 and -1 and since its to the 3rd power,
No, that's not right. sin 1 ≠ 0. Also, does (-1)3 = 1?
 
  • #10
of course, so if sinx = 0 , x = 0 and x = ∏

would that mean that sinx^3 = 0, x = [itex]\sqrt[3]{0}[/itex] and [itex]\sqrt[3]{∏}[/itex]

which means those two values are undefined

and that the values of the numerator can't equal 0 or ∏,

And since the values of the numerator equal 2/∏ and 3∏/2,

The only x value that exists within the given interval [0,∏] is x = ∏/2

Is this correct?
 
  • #11
eumyang
Homework Helper
1,347
10
of course, so if sinx = 0 , x = 0 and x = ∏

would that mean that sinx^3 = 0, x = [itex]\sqrt[3]{0}[/itex] and [itex]\sqrt[3]{∏}[/itex]
No, that's not right either. You're getting confused with regards to the cube root. From
[itex]\sin^3 x = 0[/itex]
take the cube root of both sides first.
[itex]\sqrt[3]{\sin^3 x} = \sqrt[3]{0}[/itex]
[itex]\sin x = 0[/itex]
Then find x, which you did (0 and π).
The rest looks okay to me.
 
  • #12
Yea, I was confused between getting the values x = (0, ∏) and x = (^3√0, ^3√ ∏)

This was my guess that lead me to the incorrect answer

[itex]sin^{3}x = 0[/itex]

[itex]arcsin^{3}0 = ∏[/itex]

[itex]^{3}√[/itex][itex]arcsin^{3}0[/itex] = [itex]^{3}√∏[/itex]

arcsin0 = [itex]^{3}√∏[/itex]

sin([itex]^{3}√∏) = 0[/itex]

.....and the same process for the other x value of 0.

But now I realize that [itex]sin^{3}√∏[/itex] can't equal 0

Because it is sin∏ that equals 0!!

So now I believe I understand why its correct to take the cube root of both sides first as you said.
 
Last edited:

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