# rolle's theorem

• December 21st 2009, 01:00 PM
wisezeta
rolle's theorem
for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]
• December 21st 2009, 01:06 PM
skeeter
Quote:

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.
• December 21st 2009, 04:29 PM
airportman92
Quote:

Originally Posted by skeeter
try instead the Intermediate Value Theorem.

can u do it using rolle's theorem?
• December 21st 2009, 04:42 PM
skeeter
Quote:

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.
• December 21st 2009, 04:43 PM
airportman92
Quote:

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
• December 21st 2009, 04:59 PM
skeeter
Quote:

Originally Posted by airportman92
how would you do it using rolle's theorem though just for fun

it would not be fun.
• December 21st 2009, 05:03 PM
airportman92
Quote:

Originally Posted by skeeter
it would not be fun.

but how do u do it hah
• December 21st 2009, 05:05 PM
skeeter
Quote:

Originally Posted by airportman92
but how do u do it hah

I'm not sure.
• December 21st 2009, 05:25 PM
airportman92
someone how do you use rolles theorem to do this one lol
• December 21st 2009, 06:04 PM
Jose27
It's a useless detour (at least how I did it is), but since you want it so badly:

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

As skeeter said, not pretty at all.