Problem 33 - Essential Calculus

Question

Determine whether the Mean Value Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If the Mean Value Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). $$ f(x)=\sqrt{2-x},[-7,2] $$

Step-by-Step Solution

Answer

The Mean Value Theorem can be applied to the function \(f(x) = \sqrt{2-x}\) on the interval \([-7,2]\). There is one value of \(c\) in the open interval \((-7,2)\) that satisfies the equation \(f'(c) = \frac{f(2) - f(-7)}{2 - -7} \), which is \(c = 1\).\n

Step 1: Find \(f'(x)\)

The derivative of \(f(x) = \sqrt{2-x}\) is given by \(f'(x) = - \frac{1}{2\sqrt{2-x}}\).

Step 2: Check Continuity and Differentiability

The function \(f(x) = \sqrt{2-x}\) is continuous on the interval \([-7,2]\) because the square root function is continuous for its domain and \(2-x\) is greater than or equal to 0 for all \(x\) in \([-7,2]\). The function \(f(x) = \sqrt{2-x}\) is differentiable on the interval \((-7,2)\) because the square root function is differentiable for its domain apart from the endpoint and \(2-x\) is greater than 0 for all \(x\) in \((-7,2)\). As function \(f(x)\) is both continuous on the closed interval \([-7,2]\) and differentiable on the open interval \((-7,2)\), the Mean Value Theorem can be applied.

Step 3: Find the Value of \(c\)

For the Mean Value theorem, we need to solve the equation \(f'(c) = \frac{f(2) - f(-7)}{2 - -7}\). Substituting the values into the MVT equation, we get \(- \frac{1}{2\sqrt{2-c}} = \frac{\sqrt{2-2} - \sqrt{2-(-7)}}{2 - -7} = \frac{- \sqrt{9}}{9} = -\frac{1}{3}\). To solve this for \(c\), we can cross multiply and simplify, to get the equation \(-1 = 3\sqrt{2-c}\). Squaring both sides turn this into 1 = 9(2 - c) and thus \(c = 1\). Therefore, there is one number in the interval \((-7,2)\) such that the derivative of the function at that point equals the average rate of change of the function on the interval.

Step 4: Check if Value is Within the Interval

The computed \(c = 1\) is within the interval \((-7,2)\), thus the Mean Value Theorem could be applied and the value of \(c\) such that \(f'(c) = \frac{f(2) - f(-7)}{2 - (-7)}\) is \(c = 1\).

Key Concepts

Continuity

Continuity is a fundamental concept in calculus that ensures a function behaves smoothly without any breaks or jumps within a given interval. For the Mean Value Theorem to apply, a function must be continuous on a closed interval \([a, b]\). In our exercise, the function \(f(x) = \sqrt{2-x}\) is examined over the interval \([-7, 2]\).

This function is continuous because square root functions are inherently continuous within their domain. Specifically, as long as the expression under the square root remains non-negative, you can expect continuity:

• For \(f(x) = \sqrt{2-x}\), the expression \(2-x\) is always non-negative in the interval \([-7, 2]\).
• Thus, the function is smooth and unbroken throughout this interval.

Recognizing continuity helps pave the way for other calculus applications, as it confirms the function's well-behaved nature across a given stretch.

Differentiability

Differentiability extends the idea of continuity, requiring that the function not only be smooth but also have a defined slope at each point within the specified interval. For the Mean Value Theorem, the function must be differentiable in the open interval \((a, b)\).

The function \(f(x) = \sqrt{2-x}\) is differentiable in the interval \((-7, 2)\) because:

• The square root function is differentiable wherever the expression under the root is positive.
• In the interval \((-7, 2)\), \(2-x > 0\), ensuring differentiability across the entire open interval.

It's important to note that differentiability implies continuity but not vice versa. So, checking differentiability confirms the function doesn’t have sharp turns or cusps, allowing you to find the slope at each point (except the endpoints here).

Derivative

The derivative is a core concept that measures how a function changes as its input changes. It represents the slope of the tangent line to the function at any given point. For the Mean Value Theorem, the derivative helps identify specific points where the function's instantaneous rate of change equals its average rate of change over the interval.

For our function, \(f(x) = \sqrt{2-x}\), the derivative is found as follows:

• The derivative \(f'(x) = -\frac{1}{2\sqrt{2-x}}\).
• This equation tells us how steep the function is at any point within the interval.

In the Mean Value Theorem application, you solve for \(c\) where \(f'(c)\) matches the average rate of change (provided by \(\frac{f(b) - f(a)}{b-a}\)). For this problem, \(c=1\) within the interval, indicating a point where these rates coincide, confirming practical understanding and completion of the theorem’s application.