1. ## Rolles Theorem

Determine if the Rolles theorem can be applied. If it applies find "c" such that f(x)=(x-2)(x+3)^2, [-3,2]

2. Originally Posted by sderosa518
Determine if the Rolles theorem can be applied. If it applies find "c" such that f(x)=(x-2)(x+3)^2, [-3,2]
f is differentiable and continuous since its a polynomial function
and $f(-3)=f(2)$
so the rolle's theorem can be applied here

I do not know what do you mean with:

Originally Posted by sderosa518
find "c" such that f(x)=(x-2)(x+3)^2, [-3,2]
I think you mean find "c" that satisfies the rolle's theorem, right?

3. Yes!!!

4. Originally Posted by sderosa518
Yes!!!
you want to solve $f'(c)=0$ for c ..
so ?

5. This is what I got. I use the product rule to solve.

3x^2+8x-3

Not sure if I am right or not

6. Originally Posted by sderosa518
This is what I got. I use the product rule to solve.

3x^2+8x-3

Not sure if I am right or not
correct
so $f'(x)=3x^2+8x-3$
hence $f'(c)=3c^2+8c-3$
$f'(c)=0 \rightarrow 3c^2+8c-3=0$
now solve the last equation for c

7. Any hint on how to solve it?

Should I factor a 3 out or factoriztion (x-1)(x+1)
or just simply solve for zero

8. Originally Posted by sderosa518
Any hint on how to solve it?

Should I factor a 3 out or factoriztion (x-1)(x+1)
or just simply solve for zero
you should know how to solve it

9. finale answer is 2, -14/3

Correct???

10. wrong
you should get -3 and 1/3

11. Curious how did you get?

12. Originally Posted by sderosa518
Curious how did you get?

I will be happy if i see your complete work

13. F(x)= 3x^2+8x-3

-8 -+Sqrt(8^2-4(3)(-3))
------------------------
2(3)

-8 +-sqrt(100)
--------------
6

-8+-10
------- = 2, -14/3
6

Do you see where I got my answer. When you or if you show mean, I am going to feel stupid!??!

14. Originally Posted by sderosa518
F(x)= 3x^2+8x-3

-8 -+Sqrt(8^2-4(3)(-3))
------------------------
2(3)

-8 +-sqrt(100)
--------------
6

-8+-10
------- = 2, -14/3
6

Do you see where I got my answer. When you or if you show mean, I am going to feel stupid!??!
$\frac{-8-10}{6}$ and $\frac{-8+10}{6}$