Question

Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem. (Enter your answers as a comma-separated list.) f(x) = x3 − x2 − 2x + 1, [0, 2]

Answer 1

Answer:

$c=1.215$

Step-by-step explanation:

Rolle's theorem states that if f is a continuous function defined on a closed interval [a, b] differentiable on the open interval (a, b) and f (a) = f (b), then:

There is at least one point c in the open interval (a, b) such that f '(c) = 0

Given a function:

$f(x)=x^3-x^2-2x+1$

and a interval:

$[0,2]$

Is f(x) continuous over [0,2]?

Yes, it is, because the domain of this function is $(-\infty, \infty)$

Is f'(x) differentiable on the open interval (0,2) ?

First, let's find f'(x):

$f'(x)=3x^2-2x-2$

Therefore f'(x) is differentiable on the open interval (0,2) because the domain of f'(x) is $(-\infty, \infty)$

Is f(0)=f(2) ?

$f(0)=0^3-0^2-2(0)+1=1$

$f(2)=2^3-2^2-2(2)+1=8-4-4+1=1$

Hence:

$f(0)=f(2)=1$

Now, we have verified that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. So, let's find all numbers c that satisfy the conclusion of Rolle's Theorem:

$f'(x)=0\\\\3x^2-2x-2=0$

Let's find the roots using the quadratic equation:

$x=\frac{-(-2) \pm\sqrt{(-2^2)-4(3)(-2)} }{2(3)} =\frac{2\pm \sqrt{4+24} }{6} =\frac{2\pm 2\sqrt{7} }{6} \\\\Hence\\\\x=\frac{1}{3} +\frac{\sqrt{7} }{3}\approx1.215 \\\\or\\\\x=\frac{1}{3} -\frac{\sqrt{7} }{3} \approx -0.549$

Since:

$1.215\in (0,2)\\\\-0.549 \notin (0,2)$

The value of c that satisfies the conclusion of Rolle's Theorem is:

$c=1.215$