4. find the value c that satisfies rolles theorem f(x)=x³ - 2x² - x - 1 on -1,2 rolles theorem doesnt apply…

Explanation:

Step1: Check Rolle's Theorem conditions

1. $f(x)=x^{3}-2x^{2}-x - 1$ is a polynomial, so it is continuous on $[-1,2]$.
2. The derivative $f'(x)=3x^{2}-4x - 1$ exists for all $x$, so $f(x)$ is differentiable on $(-1,2)$.
3. Calculate $f(-1)=(-1)^{3}-2(-1)^{2}-(-1)-1=-1 - 2 + 1-1=-3$ and $f(2)=2^{3}-2\times2^{2}-2 - 1=8 - 8-2 - 1=-3$. Since $f(-1)=f(2)$, Rolle's Theorem applies.

Step2: Set $f'(c) = 0$

Set $3c^{2}-4c - 1=0$. Using the quadratic - formula for a quadratic equation $ax^{2}+bx + c = 0$ ($a = 3$, $b=-4$, $c=-1$), where $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. So $c=\frac{4\pm\sqrt{(-4)^{2}-4\times3\times(-1)}}{2\times3}=\frac{4\pm\sqrt{16 + 12}}{6}=\frac{4\pm\sqrt{28}}{6}=\frac{4\pm2\sqrt{7}}{6}=\frac{2\pm\sqrt{7}}{3}$, and both values $\frac{2+\sqrt{7}}{3}\approx1.55$ and $\frac{2 - \sqrt{7}}{3}\approx - 0.22$ are in the interval $(-1,2)$.

Answer:

$\frac{2\pm\sqrt{7}}{3}$