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

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- Dec 21st 2009, 01:00 PMwisezetarolle's theorem
for g(x)=x^3 + 3x^2 -2, show that there must ne one real root on [0,2]

- Dec 21st 2009, 01:06 PMskeeter
- Dec 21st 2009, 04:29 PMairportman92
- Dec 21st 2009, 04:42 PMskeeter
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. - Dec 21st 2009, 04:43 PMairportman92
- Dec 21st 2009, 04:59 PMskeeter
- Dec 21st 2009, 05:03 PMairportman92
- Dec 21st 2009, 05:05 PMskeeter
- Dec 21st 2009, 05:25 PMairportman92
someone how do you use rolles theorem to do this one lol

- Dec 21st 2009, 06:04 PMJose27
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.