Question

determine whether the mean value theorem can be applied to f on the closed interval a, b. (select all that apply.)

f(x) = \sqrt{3 - x}, -13, 3

yes, the mean value theorem can be applied.

no, f is not continuous on a, b.

no, f is not differentiable on (a, b).

none of the above.

if the mean value theorem can be applied, find all values of c in the open interval (a, b) such that f(c) = \frac{f(b) - f(a)}{b - a}. (enter your answers as a comma - separated list. if the mean value theorem cannot be applied, enter na.)

c =

Explanation:

Step1: Check continuity

The function \(y = \sqrt{3 - x}\) is a square - root function. The domain of \(y=\sqrt{3 - x}\) is \(3-x\geq0\) or \(x\leq3\). Since the interval \([- 13,3]\) is within the domain of \(y = \sqrt{3 - x}\), and square - root functions are continuous on their domains, \(f(x)\) is continuous on \([-13,3]\).

Step2: Check differentiability

Differentiate \(f(x)=\sqrt{3 - x}=(3 - x)^{\frac{1}{2}}\) using the chain rule. The derivative \(f^{\prime}(x)=\frac{1}{2}(3 - x)^{-\frac{1}{2}}\times(-1)=-\frac{1}{2\sqrt{3 - x}}\). The derivative exists for all \(x\in(-13,3)\) (because when \(x = 3\), the derivative is not defined as we have a division by zero in \(f^{\prime}(x)\), but the Mean Value Theorem requires differentiability on the open interval \((a,b)\)).

Step3: Apply the Mean Value Theorem formula

First, find \(f(a)\) and \(f(b)\). When \(a=-13\), \(f(-13)=\sqrt{3-(-13)}=\sqrt{16} = 4\). When \(b = 3\), \(f(3)=\sqrt{3 - 3}=0\).
The formula \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\). Substitute \(a=-13\), \(b = 3\), \(f(a) = 4\), \(f(b)=0\) into \(\frac{f(b)-f(a)}{b - a}\): \(\frac{0 - 4}{3-(-13)}=\frac{-4}{16}=-\frac{1}{4}\).
Set \(f^{\prime}(c)=-\frac{1}{4}\), so \(-\frac{1}{2\sqrt{3 - c}}=-\frac{1}{4}\).
Cross - multiply: \(2\sqrt{3 - c}=4\).
Divide both sides by 2: \(\sqrt{3 - c}=2\).
Square both sides: \(3 - c = 4\).
Solve for \(c\): \(c=-1\).

Answer:

Yes, the Mean Value Theorem can be applied.
\(c=-1\)