• Oct 10th 2013, 07:49 PM
rhcprule3
Let F(x)=f(f(x)) and G(x)=F(x)^2 . You also know that f(8)=2, f(2)=3, f ' (2)=11, f '(8)=3

Find and .

I have no idea of where to even start on this
• Oct 10th 2013, 08:10 PM
SlipEternal
Use the chain rule. If \$\displaystyle c(x) = a(b(x))\$ for some differentiable functions \$\displaystyle a,b\$, then \$\displaystyle c'(x) = a'(b(x))b'(x)\$ by the chain rule. Do something similar to find \$\displaystyle F'(x)\$ and \$\displaystyle G'(x)\$.
• Oct 11th 2013, 03:49 AM
rhcprule3
ok so how do I set up the function to find F(x)? Is it something along the lines of F(f(x), so I would use F(f(2))=h(x)?
• Oct 11th 2013, 03:59 AM
Plato
Quote:

Originally Posted by rhcprule3
Let F(x)=f(f(x)) and G(x)=F(x)^2 . You also know that f(8)=2, f(2)=3, f ' (2)=11, f '(8)=3
Find Fhttp://webwork.morris.umn.edu/webwor...100/char30.png(8)= and Ghttp://webwork.morris.umn.edu/webwor...100/char30.png(8)=.

Quote:

Originally Posted by rhcprule3
ok so how do I set up the function to find F(x)? Is it something along the lines of F(f(x), so I would use F(f(2))=h(x)?

\$\displaystyle F'(x)=f'(f(x))f'(x)~\&~G'(x)=2F(x)F'(x)\$

Now you are give all necessary information to complete the substitutions.
• Oct 11th 2013, 04:25 AM
rhcprule3
ok but what do I do with the given functions? like for f(8)=2, do I just plug in f(8) for f(x)?
• Oct 11th 2013, 04:27 AM
rhcprule3
Like 2(f(8)8(f(3)? and i have no clue how to get G'(x). Id rather be walked through the problem so I can learn step by step how to do this problem...
• Oct 11th 2013, 04:38 AM
Plato
Quote:

Originally Posted by rhcprule3
Like 2(f(8)8(f(3)? and i have no clue how to get G'(x). Id rather be walked through the problem so I can learn step by step how to do this problem...

Well, it is evident to me that your misunderstandings are so profound that the help you need is beyond what we can give. You need to sit down with a live instructor and discuss what you can do to strengthen your understanding of the background material necessary for being successful in this course.
• Oct 11th 2013, 05:27 AM
SlipEternal