Critical Points and Absolute Extrema

[meks]

Here is a link to the problem.

http://i457.photobucket.com/albums/qq29 ... umber5.jpg

 

[Deleted member 4993]

Here is a link to the problem.

http://i457.photobucket.com/albums/qq29 ... umber5.jpg

Please show us your work, and exactly where you are stuck - so that we know where to begin to help you.

 

[meks]

Here's the problem

http://i457.photobucket.com/albums/qq29 ... umber5.jpg

Here's what I've done so far

\(\displaystyle f(x) = ax^3 + bx\)


\(\displaystyle f'(x) = 3ax^2 + b\)


\(\displaystyle f''(x) = 6ax\)


\(\displaystyle \kappa(x) = \frac{6ax}{\left[1 + (3ax^2 + b)^2\right]^{\frac{3}{2}}}\)


\(\displaystyle \kappa'(x)=\frac{(1+(3ax^2+b)^2)^{\frac{3}{2}}](6a)-(6ax)(\frac{3}{2})(1+(3ax^2+b)^2)^{\frac{1}{2}}(2)(3ax^2+b)(6ax)}{(1+(3ax^2+b)^2)^3}\)

I need help with the critical points. I know they are when \(\displaystyle \kappa'(x)=0\). But how would I further simplify the equation?

Thanks

 

[stapel]

For those who may not be able to view the picture of the exercise, the text of the exercise is as follows:

Let \(\displaystyle a\) and \(\displaystyle b\) be constants with \(\displaystyle a\, >\, 0\). Let \(\displaystyle f\,:\,\mathbb{R}\,\rightarrow\, \mathbb{R}\) be defined by \(\displaystyle f(x)\, =\, ax^3\, +\, bx\). Let \(\displaystyle \kappa (x)\) be as shown below:

. . . . .\(\displaystyle \kappa (x)\, :=\, \frac{f^{''}(x)}{\left(1\, +\, \left(f^'(x)\right)^2\right)^{\frac{3}{2}}}\)

Find the critical points of \(\displaystyle \kappa\) and use the first derivative test to classify them. What are the absolute extrema of \(\displaystyle \kappa\)? Would you want to use the second derivative test here? Why or why not?

 

[Deleted member 4993]

Here's the problem

http://i457.photobucket.com/albums/qq29 ... umber5.jpg

Here's what I've done so far

\(\displaystyle f(x) = ax^3 + bx\)


\(\displaystyle f'(x) = 3ax^2 + b\)


\(\displaystyle f''(x) = 6ax\)


\(\displaystyle \kappa(x) = \frac{6ax}{\left[1 + (3ax^2 + b)^2\right]^{\frac{3}{2}}}\)


\(\displaystyle \kappa'(x)=\frac{(1+(3ax^2+b)^2)^{\frac{3}{2}}](6a)-(6ax)(\frac{3}{2})(1+(3ax^2+b)^2)^{\frac{1}{2}}(2)(3ax^2+b)(6ax)}{(1+(3ax^2+b)^2)^3}\)

First - simplify the numerator a bit by multiplying stuff out.

Then factor out [1 + (3ax[sup:x526esiw]2[/sup:x526esiw] +b) [sup:x526esiw]2[/sup:x526esiw]][sup:x526esiw]1/2[/sup:x526esiw] - and simplify further

I need help with the critical points. I know they are when \(\displaystyle \kappa'(x)=0\). But how would I further simplify the equation?

Thanks