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

• LearninDaMath
Thanks for the help.In summary, the function f(x) = 2x + cotx over the interval [0,pi] has increasing intervals at x = pi/4 and 3pi/4 and decreasing intervals at x = 0 and pi. The concavity of the function is positive at x = pi/4 and 3pi/4 and negative at x = 0 and pi. To find the inflection points, the second derivative is set equal to 0. By taking the cube root of both sides and finding the values that make the numerator and denominator equal to 0, the only valid inflection point within the interval is x = pi/2.

## 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...

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.

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.

LearninDaMath said:
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.

$$\frac{2\cos x}{\sin^3 x}$$

You are trying to differentiate this right?

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

$$2 (\frac{\cos x}{\sin x^3}$$

Hint: $\frac{a}{b^2} = \frac{a}{b}\frac{1}{b} or \frac{1}{x} = x^{-1}$

There is also a simple differentiation rule that is very helpful in calculus I. Rings a bell?

Nano-Passion said:
$$\frac{2\cos x}{\sin^3 x}$$

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

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.

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 ($\sqrt{1}$ and$\sqrt{-1}$)

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?

Wait a minute...

2 should not be included in setting cosx = 0

because 2 can be made a separate 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?

LearninDaMath said:
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?

of course, so if sinx = 0 , x = 0 and x = ∏

would that mean that sinx^3 = 0, x = $\sqrt{0}$ and $\sqrt{∏}$

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?

LearninDaMath said:
of course, so if sinx = 0 , x = 0 and x = ∏

would that mean that sinx^3 = 0, x = $\sqrt{0}$ and $\sqrt{∏}$
No, that's not right either. You're getting confused with regards to the cube root. From
$\sin^3 x = 0$
take the cube root of both sides first.
$\sqrt{\sin^3 x} = \sqrt{0}$
$\sin x = 0$
Then find x, which you did (0 and π).
The rest looks okay to me.

Yea, I was confused between getting the values x = (0, ∏) and x = (^3√0, ^3√ ∏)

$sin^{3}x = 0$

$arcsin^{3}0 = ∏$

$^{3}√$$arcsin^{3}0$ = $^{3}√∏$

arcsin0 = $^{3}√∏$

sin($^{3}√∏) = 0$

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

But now I realize that $sin^{3}√∏$ 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: