Rolle's Theorem Problem

[Jason76]

\(\displaystyle f(x) = x^{3} - x^{2} - 2x + 9\)

It is a polynomial so continuous and differentiable across all real numbers.

\(\displaystyle [0,2]\)

\(\displaystyle f(0) = (0)^{3} - (0)^{2} - 2(0) + 9 = 9\)

\(\displaystyle f(2) = (2)^{3} - (2)^{2} - 2(2) + 9 = 9\)

\(\displaystyle f'(x) = 3x^{2} - 2x - 2\)

\(\displaystyle 3x^{2} - 2x - 2 = 0\)


\(\displaystyle \dfrac{-(-2) \pm \sqrt{(-2)^{2} - 4(3)(-2)}}{2(3)}\)

\(\displaystyle \dfrac{2 \pm \sqrt{(4 -(-24)}}{6}\)

\(\displaystyle \dfrac{2 \pm \sqrt{(28)}}{6}\) ?? This heading in wrong direction.

 

[stapel]

\(\displaystyle \dfrac{2 \pm \sqrt{(28)}}{6}\) ?? This heading in wrong direction.

Dunno. What question are you supposed to be answering? In what direction are you intended to be heading?

 

[Jason76]

Dunno. What question are you supposed to be answering? In what direction are you intended to be heading?

The square root of 28 is a strange number, so it doesn't work well with the quadratic formula. So something is wrong somewhere.

 

[Deleted member 4993]

We still don't know - what the question was!!

Just because you got an irrational number - does not mean that you are headed wrong way.

There "more" irrational numbers than there are rational numbers (and people too...)!!

 

[Jason76]

A situation of not going thru with the problem all the way. The square root of 28 had to be pushed thru. So the final answer comes out to be \(\displaystyle \dfrac{1 + \sqrt{7}}{3}\)