Finding critical numbers

[stinajeana]

Find every critical number of f(x)=ln(x^2-1)+8/x^2-1

So I know that in order to begin I must find the derivative of it all but I don't know what to do afterwards. Any help would be great :wink:

 

[pka]

Find every critical number of f(x)=ln(x^2-1)+8/x^2-1

So I know that in order to begin I must find the derivative of it all but I don't know what to do afterwards. Any help would be great

Is the function \(\displaystyle f(x)=\ln(x^2-1)+\dfrac{8}{x^2-1}~?\)

If so, then solve \(\displaystyle f'(x)=0~\&~f"(x)=0\)

 

[HallsofIvy]

A "critical point" of a function f(x) is given by a value of x such that either f'(x)= 0 or the derivative does not exist. It is NOT necessary to look at the second derivative- if f'(x) is not 0, there is no critical point at x, even if f''(x)= 0. For example, f(x)= x^3- 2x+ 1 has a critical point at \(\displaystyle \sqrt{2/3}\) because its derivative, 3x^2- 2 is 0 there. f''(x)= 6x is 0 when x= 0 but x= 0 does not give a "critical point".

 

[stinajeana]

A "critical point" of a function f(x) is given by a value of x such that either f'(x)= 0 or the derivative does not exist. It is NOT necessary to look at the second derivative- if f'(x) is not 0, there is no critical point at x, even if f''(x)= 0. For example, f(x)= x^3- 2x+ 1 has a critical point at \(\displaystyle \sqrt{2/3}\) because its derivative, 3x^2- 2 is 0 there. f''(x)= 6x is 0 when x= 0 but x= 0 does not give a "critical point".

Thank you!!!!

 

[lookagain]

Find > > > every critical number < < < of f(x)=ln(x^2-1)+8/x^2-1

So I know that in order to begin I must find the derivative of it all but I don't know what to do afterwards. Any help would be great :wink:

After reading these posts, there seems to be a disagreement of whether "every critical number" means just first order critical numbers (from the first derivative) or both first and second order critical numbers (from the first and second derivatives, respectively).

 

[whiteti]

Is the function \(\displaystyle f(x)=\ln(x^2-1)+\dfrac{8}{x^2-1}~?\)

If so, then solve \(\displaystyle f'(x)=0~\&~f"(x)=0\)

Hi, Im working on this problem too. and Im stuck on finding the derivative of the original equation. Help?

 

[DrPhil]

Hi, Im working on this problem too. and Im stuck on finding the derivative of the original equation. Help?

\(\displaystyle f(x) = \ln{(x^2 - 1)} + \dfrac{8}{x^2 - 1}\)

The first term is a logarithm: the derivative of ln(u) is 1/u, and applying the chain rule du/dx = 2x.

The second term I would write as \(\displaystyle 8(x^2 - 1)^{-1}\). Use the power rule followed by the chain rule:
......8/(x^2 - 1)^2 * 2x

Putting then together, \(\displaystyle \displaystyle f'(x) = \dfrac{2x}{x^2 - 1} - \dfrac{16 x}{(x^2 - 1)^2}\)

which you can simplify.

 

[whiteti]

\(\displaystyle f(x) = \ln{(x^2 - 1)} + \dfrac{8}{x^2 - 1}\)

The first term is a logarithm: the derivative of ln(u) is 1/u, and applying the chain rule du/dx = 2x.

The second term I would write as \(\displaystyle 8(x^2 - 1)^{-1}\). Use the power rule followed by the chain rule:
......8/(x^2 - 1)^2 * 2x

Putting then together, \(\displaystyle \displaystyle f'(x) = \dfrac{2x}{x^2 - 1} - \dfrac{16 x}{(x^2 - 1)^2}\)

which you can simplify.

how do i simplify? By taking another derivative?

 

[DrPhil]

how do i simplify? By taking another derivative?

Combine the fractions over a common denominator, so you can find where the numerator is 0.

EDIT: what is the domain of x?

 

[JeffM]

how do i simplify? By taking another derivative?

You are trying to find where the derivative that you have already found equals zero. How does taking the second derivative do that?