1. ## rolle's theorem

for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]

2. Originally Posted by wisezeta
for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]
try instead the Intermediate Value Theorem.

3. Originally Posted by skeeter
try instead the Intermediate Value Theorem.
can u do it using rolle's theorem?

4. Originally Posted by airportman92
can u do it using rolle's theorem?
Rolle's theorem says that if a function is continuous on [a,b], differentiable on (a,b), and f(a) = f(b) , then there is at least one value c in the interval (a,b) where f'(c) = 0.

I'm not saying it can't be done, but it would not be as simple as using the IVT.

5. Originally Posted by skeeter
Rolle's theorem says that if a function is continuous on [a,b], differentiable on (a,b), and f(a) = f(b) , then there is at least one value c in the interval (a,b) where f'(c) = 0.

I'm not saying it can't be done, but it would not be as simple as using the IVT.

how would you do it using rolle's theorem though just for fun

6. Originally Posted by airportman92
how would you do it using rolle's theorem though just for fun
it would not be fun.

7. Originally Posted by skeeter
it would not be fun.
but how do u do it hah

8. Originally Posted by airportman92
but how do u do it hah
I'm not sure.

9. someone how do you use rolles theorem to do this one lol

10. It's a useless detour (at least how I did it is), but since you want it so badly:

Let $\displaystyle g_c(x)= \frac{x^4}{4} + x^3 -2x +c=4x(x^3+4x^2-2)+c$ Let $\displaystyle h(x)=x^3+4x^2-2$ then $\displaystyle h(0)<0$ and $\displaystyle h(2)>0$ so by the intermediate value theorem there exists $\displaystyle d\in (0,2)$ such that $\displaystyle h(d)=0$ and so $\displaystyle g_c(0)=g_c(d)=c$ and by Rolle's theorem there exists $\displaystyle y\in(0,d)$ such that $\displaystyle g'_c (y)=0$ but $\displaystyle g'_c=g$

As skeeter said, not pretty at all.