14. -/1 points details my notes scalc9 3.2.507.xp. verify that the function satisfies the three hypotheses…

Explanation:

Step1: Check continuity

Polynomial functions are continuous everywhere. Since (f(x)=x^{3}-x^{2}-20x + 2) is a polynomial, it is continuous on ([0,5]).

Step2: Check differentiability

The derivative of a polynomial exists everywhere. Differentiate (f(x)) using the power - rule. (f'(x)=3x^{2}-2x - 20). So (f(x)) is differentiable on ((0,5)).

Step3: Check (f(0)) and (f(5))

Calculate (f(0)=0^{3}-0^{2}-20\times0 + 2=2). Calculate (f(5)=5^{3}-5^{2}-20\times5 + 2=125 - 25-100 + 2=2). Since (f(0)=f(5)), the three hypotheses of Rolle's Theorem are satisfied.

Step4: Find (c)

Set (f'(c)=0). So (3c^{2}-2c - 20 = 0). Use the quadratic formula (c=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for the quadratic equation (ax^{2}+bx + c = 0). Here (a = 3), (b=-2), (c=-20). Then (c=\frac{2\pm\sqrt{(-2)^{2}-4\times3\times(-20)}}{2\times3}=\frac{2\pm\sqrt{4 + 240}}{6}=\frac{2\pm\sqrt{244}}{6}=\frac{2\pm2\sqrt{61}}{6}=\frac{1\pm\sqrt{61}}{3}). We reject (c=\frac{1-\sqrt{61}}{3}) since (\frac{1-\sqrt{61}}{3}\approx\frac{1 - 7.81}{3}\lt0). The value (c=\frac{1 + \sqrt{61}}{3}\approx\frac{1+7.81}{3}=\frac{8.81}{3}\approx2.94\in(0,5)).

Answer:

(\frac{1+\sqrt{61}}{3})