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

determine whether the mean value theorem can be applied to f on the closed interval a, b. (select all that apply.)\n\nf(x) = \\sqrt{3 - x}, -13, 3\n\nyes, the mean value theorem can be applied.\n\nno, f is not continuous on a, b.\n\nno, f is not differentiable on (a, b).\n\nnone of the above.\n\nif 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.)\n\nc =

determine whether the mean value theorem can be applied to f on the closed interval a, b. (select all that apply.)\n\nf(x) = \\sqrt{3 - x}, -13, 3\n\nyes, the mean value theorem can be applied.\n\nno, f is not continuous on a, b.\n\nno, f is not differentiable on (a, b).\n\nnone of the above.\n\nif 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.)\n\nc =

Answer

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)