Second Derivative-Relative Maxima

[Justabeginner]

Homework Statement

I have to find the relative maximum, so I'm going to set the second derivative to zero and solve. But, this is where I'm stuck.

Homework Equations

f"(x)= (x^2-4)(sin(2x)) - (2x)(1+cos^2(x))/ ((1+cos^2(x))^2) [The entire equation is divided by the ((1+cos^2(x))^2)]

The Attempt at a Solution

(x^2-4)(sin(2x)) - (2x) (1+cos^2(x))= 0
(x^2-4) (sin(2x)) = (2x) (1+cos^2(x))

I feel like an idiot, because I can't get past this point ^. Any help is appreciated. Thank you!

 

[Ray Vickson]

Homework Statement

I have to find the relative maximum, so I'm going to set the second derivative to zero and solve. But, this is where I'm stuck.

Homework Equations

f"(x)= (x^2-4)(sin(2x)) - (2x)(1+cos^2(x))/ ((1+cos^2(x))^2) [The entire equation is divided by the ((1+cos^2(x))^2)]

The Attempt at a Solution

(x^2-4)(sin(2x)) - (2x) (1+cos^2(x))= 0
(x^2-4) (sin(2x)) = (2x) (1+cos^2(x))

I feel like an idiot, because I can't get past this point ^. Any help is appreciated. Thank you!

First: what is your f(x)? Second: why are you trying to find the inflection points of the graph y = f(x)?

 

[Justabeginner]

f(x) is the integral of x^2-4/((1+cos^2(x))

I am trying to find the maximum, not the inflection points, but I just realized that solving for the 2nd derivative gives you the inflection points? How would I get the maximum then?

If I plug into the second derivative, with the critical points given by the first derivative, then I get zero. This means that neither -2 or 2 are relative maxima?

 

[LCKurtz]

f(x) is the integral of x^2-4/((1+cos^2(x))

I am trying to find the maximum, not the inflection points, but I just realized that solving for the 2nd derivative gives you the inflection points? How would I get the maximum then?

If I plug into the second derivative, with the critical points given by the first derivative, then I get zero. This means that neither -2 or 2 are relative maxima?

So now you have told us that$$
f(x) = \int \frac {x^2-4}{1+\cos^2 x}\, dx$$and apparently you understand that$$
f'(x) = \frac {x^2-4}{1+\cos^2 x}$$and you have figured out that the critical points are $x=\pm 2$. I didn't check your second derivative, but you shouldn't need it anyway. Write your first derivative as$$
\frac {(x+2)(x-2)}{1+\cos^2 x}$$The denominator is always positive. Look at the signs of the factors in the numerator to see when the slope is positive or negative and use that to determine the max/min points.

 

[Justabeginner]

That's where I'm lost. They both give you zero. Are they maximums on the interval? When x < -2 and x > 2 , f(x) > 0

 

[LCKurtz]

That's where I'm lost. They both give you zero. Are they maximums on the interval? When x < -2 and x > 2 , f(x) > 0

That isn't $f(x)$ we are talking about the signs for, it is $f'(x)$. What do the signs tell you? And what about between $-2$ and $2$?

 

[Justabeginner]

I think I realize what I did wrong.

The intervals I needed to test were: x < -2 , - 2 < x < 2, and x > 2

For x < -2, f'(x) > 0
For -2 < x < 2, f'(x) < 0
For x > 2, f'(x) > 0

I come to the conclusion that (-2, 0) and (2, 0) are the relative minima and (0, -2) is the relative maximum. Is this correct?

 

[LCKurtz]

I think I realize what I did wrong.

The intervals I needed to test were: x < -2 , - 2 < x < 2, and x > 2

For x < -2, f'(x) > 0
For -2 < x < 2, f'(x) < 0
For x > 2, f'(x) > 0

I come to the conclusion that (-2, 0) and (2, 0) are the relative minima and (0, -2) is the relative maximum. Is this correct?

Maximum and minimum of what? You keep omitting what you are finding the maximum and minimum of. And therefore confusing f(x) and f'(x). Where did you get those values for the y coordinate? What are they the y values of?