Comprehensive Study Notes on the Mean Value Theorem

Fundamentals of the Mean Value Theorem

The Mean Value Theorem (MVT), or مبرهنة القيمة المتوسطة, is a foundational principle in calculus that bridges the relationship between the average rate of change of a function over an interval and the instantaneous rate of change at a specific point within that interval. For a function $f(x)$ defined on a closed interval $[a, b]$, the theorem asserts that if the function satisfies two primary conditions—continuity on the closed interval $[a, b]$ and differentiability on the open interval $(a, b)$—then there exists at least one value $c$ in the open interval $(a, b)$ such that the derivative of the function at $c$ equals the slope of the secant line connecting the endpoints of the interval. Mathematically, this is expressed as $f'(c) = \frac{f(b) - f(a)}{b - a}$. This cumulative exam focuses on applying these conditions and solving for the unknown constant $c$ across various function types, including polynomials, rational functions, and radical functions.

Application of MVT to Polynomial Functions: Question One

The first problem examines the function $f(x) = (x - 1)^4$ on the interval $[1, 3]$. As a polynomial function, it is inherently continuous on all real numbers, thereby satisfying the first condition of continuity on the closed interval $[1, 3]$. Furthermore, it is differentiable on the open interval $(1, 3)$. To find the value of $c$, we first determine the derivative of the function using the power rule and the chain rule: $f'(x) = 4(x - 1)^3$. Consequently, the instantaneous rate of change at $c$ is $f'(c) = 4(c - 1)^3$.

Next, we calculate the average rate of change (the slope of the secant line) between the endpoints $a = 1$ and $b = 3$. Evaluating the function at the endpoints yields $f(1) = (1 - 1)^4 = 0$ and $f(3) = (3 - 1)^4 = 2^4 = 16$. The average rate of change is calculated as $\frac{16 - 0}{3 - 1} = \frac{16}{2} = 8$. By equating the instantaneous rate to the average rate, we obtain the equation $4(c - 1)^3 = 8$. Dividing both sides by $4$ gives $(c - 1)^3 = 2$. Taking the cube root of both sides leads to c - 1 = \root 3 \nof 2, which simplifies to c = 1 + \root 3 \nof 2. Approximately, this value is $c \thickapprox 2.26$, which falls within the open interval $(1, 3)$, confirming the theorem's validity for this function.

Analysis of Rational Functions: Question Two

Question Two presents the rational function $f(x) = 5 - \frac{4}{x}$ on the interval $[1, 4]$. For rational functions, continuity must be checked specifically at points where the denominator equals zero. Here, the function is undefined at $x = 0$. However, since $0$ does not lie within the given interval $[1, 4]$, the function is continuous on $[1, 4]$ and differentiable on $(1, 4)$. The derivative is obtained by differentiating the constant $5$ (which equals $0$) and applying the power rule to the term $-4x^{-1}$, resulting in $f'(x) = \frac{4}{x^2}$. Thus, $f'(c) = \frac{4}{c^2}$.

To find the average rate of change, we evaluate the function at the endpoints: $f(1) = 5 - \frac{4}{1} = 1$ and $f(4) = 5 - \frac{4}{4} = 4$. The average slope is $\frac{4 - 1}{4 - 1} = \frac{3}{3} = 1$. Setting the derivative equal to the average slope gives $\frac{4}{c^2} = 1$, which implies $c^2 = 4$. Solving for $c$, we find $c = 2$ or $c = -2$. We discard $c = -2$ because it does not reside within the interval $(1, 4)$. The valid solution is $c = 2$.

Differentiability Constraints in Radical Functions: Question Three

In Question Three, the function provided is f(x) = 3\root 3 \nof {(x + 1)^2} on the interval $[-2, 7]$. This is a radical function of the form $f(x) = 3(x + 1)^{\frac{2}{3}}$. While the function is continuous for all real numbers (and thus continuous on $[-2, 7]$), we must carefully examine its differentiability. The derivative is f'(x) = 3 \times \frac{2}{3}(x + 1)^{-\frac{1}{3}} = \frac{2}{\root 3 \nof {x + 1}}. The derivative is undefined when the denominator is zero, which occurs at $x = -1$.

Because the point $x = -1$ is located within the open interval $(-2, 7)$, the function is not differentiable on $(1, 7)$. Specifically, the function has a vertical tangent or a cusp at $x = -1$. Consequently, the second condition of the Mean Value Theorem is not satisfied. Therefore, the Mean Value Theorem does not apply, and we cannot determine a value for $c$ using this theorem for the specified interval.

Applying MVT to Square Root Functions: Question Four

Question Four involves the square root function f(x) = \root \nof {9 - x^2} on the interval $[-2, 2]$. The domain of this function is restricted to values where 9 - x^2 \req 0, which corresponds to $x \text{ in } [-3, 3]$. Since the interval $[-2, 2]$ is a subset of the domain, the function is continuous on $[-2, 2]$. The derivative is f'(x) = \frac{-x}{\root \nof {9 - x^2}}, which is defined for all $x \text{ in } (-3, 3)$. Thus, the function is differentiable on the open interval $(-2, 2)$.

Evaluating the endpoints, we find f(-2) = \root \nof {9 - (-2)^2} = \root \nof 5 and f(2) = \root \nof {9 - (2)^2} = \root \nof 5. The average rate of change is \frac{\root \nof 5 - \root \nof 5}{2 - (-2)} = \frac{0}{4} = 0. Note that this specific case, where $f(a) = f(b)$, is also known as Rolle's Theorem. To find $c$, we set the derivative equal to zero: \frac{-c}{\root \nof {9 - c^2}} = 0, which simplifies to $-c = 0$, or $c = 0$. Since $0 \text{ in } (-2, 2)$, the theorem is satisfied at $c = 0$.

Determing Interval Endpoints from MVT Constants: Question Five

The final problem explores an inverse scenario where we are given that the function $f(x) = x^2 - 3x - 1$ satisfies the conditions of the Mean Value Theorem on the interval $[1, b]$. We are also given that the point representing the instantaneous rate of change is $c = 2$. Our goal is to determine the unknown upper bound $b$. First, we compute the derivative: $f'(x) = 2x - 3$. Evaluating the derivative at the known constant $c = 2$ gives $f'(2) = 2(2) - 3 = 1$.

According to the MVT, this instantaneous rate must equal the average rate of change over $[1, b]$. The average rate is calculated as $\frac{f(b) - f(1)}{b - 1}$. Evaluating the function, we find $f(b) = b^2 - 3b - 1$ and $f(1) = 1^2 - 3(1) - 1 = -3$. Substituting these into the formula yields $\frac{(b^2 - 3b - 1) - (-3)}{b - 1} = \frac{b^2 - 3b + 2}{b - 1}$. The numerator can be factored into $(b - 1)(b - 2)$. Dividing by $(b - 1)$ simplifies the expression to $b - 2$. By setting this equal to the value of the derivative at $c$, we have $b - 2 = 1$. Solving for $b$, we find that $b = 3$.