Determine whether Rolle's Theorem can be applied to f on the...

Determine whether Rolle's Theorem can be applied to $f$ on the closed interval $[a, b] .$ If Rolle's Theorem can be applied, find all values of $c$ in the open interval $(a, b)$ such that $f^{\prime}(c)=0 .$ If Rolle's Theorem cannot be applied, explain why not. $$f(x)=\frac{x^{2}-2 x}{x+2},[-1,6]$$

Key Concepts

Differentiability on an Open Interval

Differentiability on the open interval (a, b) means that the function has a defined derivative at every point within that interval. This condition implies that the function is smooth without any sharp corners or cusps, necessary for asserting the existence of a point where the slope of the tangent (the derivative) is zero.

Endpoints Equality Condition

A crucial prerequisite for applying Rolle's Theorem is that the function values at the endpoints of the interval must be equal, i.e., f(a) = f(b). This condition guarantees that the function starts and ends at the same value, making it possible to find a stationary point where the derivative is zero somewhere in between.

Rolle's Theorem

Rolle's Theorem states that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and has equal function values at the endpoints (f(a) = f(b)), then there exists at least one point c in (a, b) where the derivative f'(c) is zero. This is a fundamental tool in understanding the behavior of functions between two points where the function attains the same value.

Continuity on a Closed Interval

For a function to meet the requirements of Rolle's Theorem, it must be continuous on the entire closed interval [a, b]. Continuity ensures there are no breaks or jumps in the function over the interval, which is especially important when dealing with rational functions where division by zero or undefined points might occur.

Transcript

00:01 So for this question, we will be determining whether roles theorem can be applied for the following function from negative 1 to 6 inclusive.

00:18 And then we will be finding all the values of c from negative 1 to 6 such that f prime of c is equal to 0.

00:29 Roles theorem can be applied when three cases are true.

00:34 First is the function continuous on the interval.

00:43 And when we have a function with a denominator, the function is actually not continuous when the denominator is equal to 0.

00:51 So we can check this by setting the denominator equal to 0.

00:57 So the function is actually not continuous at x is equal to negative 2, which is non -innerable, so we have no issues.

01:08 Next, we determine if the function's differentiable on the open interval.

01:18 We do that by taking the derivative of the function, determining if it is valid, for the interval.

01:24 So if prime of x, we're going to be using quotient rule.

01:29 The denominator, multiplied by the derivative of the top, subtract it from the numerator, and the derivative of the denominator all over the denominator squared.

01:46 There's no common factors in the numerator and denominator.

01:49 So again, all we need to check is the denominator to see if the function does not exist anywhere.

01:58 And again, the function.

02:00 Function only doesn't exist to x equals negative 2.

02:04 This is the derivative of the function.

02:07 So we know that the function's not differentiable at this value only, which is not in the interval.

02:15 So again, we're fine.

02:17 The last check is, does the function have an equal value at both a and b, which are the bounds? so is f at negative 1? is f at negative 1 equal to f at 6.

02:44 And check that just by plugging in...

Please give Ace some feedback