Mean Value Theorem Analysis Determining Functions Satisfying MVT Conditions On [-4 4]

Introduction to the Mean Value Theorem

The Mean Value Theorem (MVT) is a cornerstone of calculus, providing a vital link between the average rate of change of a function over an interval and its instantaneous rate of change at a specific point within that interval. Understanding this theorem is crucial for students and professionals alike, as it underpins many advanced mathematical concepts and practical applications. To fully grasp the MVT, it’s essential to know its conditions and how to apply them to various functions. This article delves into the conditions of the MVT, specifically focusing on determining which functions satisfy these conditions over a given interval. Let's explore the MVT in detail, highlighting its significance in calculus and its applicability in real-world scenarios.

The Mean Value Theorem states that if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists at least one point $c$ in $(a, b)$ such that the instantaneous rate of change at $c$ is equal to the average rate of change over the interval $[a, b]$. Mathematically, this is expressed as:

$$f'(c) = \frac{f(b) - f(a)}{b - a}$$

This theorem has significant implications in various fields, including physics, engineering, and economics. For instance, in physics, it can be used to relate average velocity to instantaneous velocity. In economics, it can help analyze marginal cost and average cost functions. The MVT’s versatility stems from its ability to connect local behavior (the derivative at a point) with global behavior (the average rate of change over an interval). Therefore, mastering the MVT is not only essential for calculus but also for understanding many real-world phenomena.

Before applying the MVT, it is imperative to check whether the function meets the two critical conditions: continuity on the closed interval $[a, b]$ and differentiability on the open interval $(a, b)$. Continuity ensures that the function has no breaks or jumps within the interval, meaning you can draw the graph of the function without lifting your pen. Differentiability, on the other hand, means that the function has a derivative at every point in the open interval, implying that the graph is smooth and has no sharp corners or vertical tangents. If either of these conditions is not met, the MVT cannot be applied. Understanding these conditions is vital because it dictates the applicability of the theorem and the validity of any conclusions drawn from it. Ignoring these conditions can lead to incorrect interpretations and applications of the theorem. For example, a function with a discontinuity or a sharp corner within the interval will not satisfy the MVT, and attempting to apply the theorem in such cases would yield meaningless results.

Evaluating the Functions

I. $f(x) = e^{x^2}$

The first function we need to analyze is $f(x) = e^{x^2}$. To determine if this function satisfies the conditions of the Mean Value Theorem (MVT) over the interval $[-4, 4]$, we must check for both continuity and differentiability within this interval. Continuity is a fundamental property, ensuring that there are no breaks, jumps, or asymptotes in the function’s graph. Differentiability, on the other hand, implies that the function has a well-defined derivative at every point in the interval, meaning the graph is smooth without any sharp corners or vertical tangents.

First, let's consider the continuity of $f(x) = e^{x^2}$. The function $e^{x^2}$ is a composite function, where the outer function is the exponential function $e^u$ and the inner function is $u = x^2$. Exponential functions are continuous for all real numbers, and polynomial functions like $x^2$ are also continuous for all real numbers. The composition of continuous functions is continuous, so $f(x) = e^{x^2}$ is continuous over its entire domain, which includes the interval $[-4, 4]$. This means there are no points within the interval where the function is undefined or has any discontinuities. We can confidently say that $f(x)$ meets the continuity requirement for the MVT over the interval $[-4, 4]$.

Next, we must examine the differentiability of $f(x) = e^{x^2}$ over the open interval $(-4, 4)$. To do this, we need to find the derivative of the function and check if it exists for all $x$ in the interval. The derivative of $f(x)$ can be found using the chain rule. Let $u = x^2$, then $f(x) = e^u$. The derivative of $e^u$ with respect to $u$ is $e^u$, and the derivative of $x^2$ with respect to $x$ is $2x$. Applying the chain rule, we get:

$$f'(x) = \frac{d}{dx} e^{x^2} = e^{x^2} \cdot \frac{d}{dx} (x^2) = e^{x^2} \cdot 2x = 2xe^{x^2}$$

The derivative $f'(x) = 2xe^{x^2}$ is a product of a linear function ($2x$) and an exponential function ($e^{x^2}$). Both of these functions are differentiable for all real numbers. Therefore, their product, $f'(x)$, is also differentiable for all real numbers. This means that $f'(x)$ exists for every $x$ in the open interval $(-4, 4)$. Thus, $f(x) = e^{x^2}$ satisfies the differentiability condition of the MVT over the interval $(-4, 4)$.

Since $f(x) = e^{x^2}$ is both continuous on the closed interval $[-4, 4]$ and differentiable on the open interval $(-4, 4)$, it satisfies the conditions of the Mean Value Theorem. This implies that there exists at least one point $c$ within the interval $(-4, 4)$ where the instantaneous rate of change (the derivative at $c$) is equal to the average rate of change of the function over the interval $[-4, 4]$.

II. $g(x) = 2^{1/x^2}$

Let's now examine the function $g(x) = 2^{1/x^2}$ to determine if it satisfies the Mean Value Theorem (MVT) conditions over the interval $[-4, 4]$. As with the previous function, we need to assess both continuity and differentiability. Continuity ensures that the function is defined and unbroken across the interval, while differentiability means the function has a derivative at every point in the interval, implying a smooth graph without sharp turns or vertical tangents.

To check for continuity, we first identify any potential points of discontinuity. The exponent $1/x^2$ in $g(x) = 2^{1/x^2}$ suggests a possible issue when $x = 0$, as division by zero is undefined. This means that the function $g(x)$ is not defined at $x = 0$, which lies within the interval $[-4, 4]$. Therefore, $g(x)$ is not continuous on the closed interval $[-4, 4]$ because it has a discontinuity at $x = 0$. This single point of discontinuity is enough to disqualify the function from satisfying the continuity condition of the MVT over the given interval.

Since the function $g(x)$ is not continuous on the interval $[-4, 4]$, we do not need to further investigate its differentiability. The Mean Value Theorem requires both continuity on the closed interval and differentiability on the open interval. If either of these conditions is not met, the theorem cannot be applied. In this case, the lack of continuity at $x = 0$ is sufficient to conclude that $g(x)$ does not satisfy the conditions of the MVT over the interval $[-4, 4]$.

It’s important to recognize that even if a function is differentiable at most points within an interval, the absence of continuity at even a single point within that interval invalidates the application of the MVT. This highlights the critical nature of both conditions – continuity and differentiability – in the theorem’s applicability. The presence of a discontinuity means there is a break in the function’s graph, preventing the smooth transition required for the MVT to hold.

In summary, the function $g(x) = 2^{1/x^2}$ fails to satisfy the Mean Value Theorem over the interval $[-4, 4]$ due to its discontinuity at $x = 0$. This demonstrates that a thorough check for continuity is essential before attempting to apply the MVT, as a single discontinuity can prevent the theorem from being valid.

III. $h(x) = \sqrt{x + 4}$

Now, let's analyze the function $h(x) = \sqrt{x + 4}$ to determine if it satisfies the conditions of the Mean Value Theorem (MVT) over the interval $[-4, 4]$. As with the previous functions, we need to evaluate both continuity and differentiability within this interval. Continuity ensures the function's graph is unbroken, while differentiability means the function has a derivative at every point in the open interval, indicating a smooth curve without sharp corners or vertical tangents.

First, we need to assess the continuity of $h(x) = \sqrt{x + 4}$ on the closed interval $[-4, 4]$. The square root function is continuous wherever its argument is non-negative. Thus, we need to ensure that $x + 4 \geq 0$ for all $x$ in the interval $[-4, 4]$. Solving this inequality, we find that $x \geq -4$. This means the function is defined and continuous for all $x$ greater than or equal to $-4$. Since our interval is $[-4, 4]$, the function is continuous on this interval because the argument of the square root is non-negative throughout the interval. Therefore, $h(x)$ satisfies the continuity condition for the MVT on the interval $[-4, 4]$.

Next, we examine the differentiability of $h(x) = \sqrt{x + 4}$ on the open interval $(-4, 4)$. To do this, we first find the derivative of $h(x)$. We can rewrite $h(x)$ as $(x + 4)^{1/2}$ and use the power rule and chain rule to find its derivative:

$$h'(x) = \frac{d}{dx} (x + 4)^{1/2} = \frac{1}{2}(x + 4)^{-1/2} \cdot \frac{d}{dx}(x + 4) = \frac{1}{2}(x + 4)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x + 4}}$$

Now, we need to determine where $h'(x)$ exists on the open interval $(-4, 4)$. The derivative $h'(x) = \frac{1}{2\sqrt{x + 4}}$ involves a square root in the denominator, which means we need to ensure that $x + 4 > 0$ (since we cannot divide by zero). This gives us $x > -4$. However, the open interval we are considering is $(-4, 4)$, which does not include $x = -4$. Thus, the derivative $h'(x)$ exists for all $x$ in the interval $(-4, 4)$. This means that $h(x)$ is differentiable on the open interval $(-4, 4)$.

Since $h(x) = \sqrt{x + 4}$ is both continuous on the closed interval $[-4, 4]$ and differentiable on the open interval $(-4, 4)$, it satisfies the conditions of the Mean Value Theorem. This implies that there exists at least one point $c$ in the interval $(-4, 4)$ where the instantaneous rate of change (the derivative at $c$) is equal to the average rate of change of the function over the interval $[-4, 4]$.

Conclusion

After a thorough analysis of the three functions, we can now determine which ones satisfy the conditions of the Mean Value Theorem (MVT) over the interval $[-4, 4]$. The MVT requires a function to be continuous on the closed interval and differentiable on the open interval. These conditions ensure that there exists at least one point within the interval where the instantaneous rate of change equals the average rate of change.

• Function I: $f(x) = e^{x^2}$ was found to be continuous and differentiable over the interval $[-4, 4]$. The exponential function $e^{x^2}$ is continuous for all real numbers, and its derivative, $2xe^{x^2}$, also exists for all real numbers. Therefore, $f(x)$ satisfies the conditions of the MVT.
• Function II: $g(x) = 2^{1/x^2}$ was found to be discontinuous at $x = 0$, which lies within the interval $[-4, 4]$. The function is undefined at this point due to division by zero in the exponent. Since it is not continuous on the closed interval, $g(x)$ does not satisfy the conditions of the MVT.
• Function III: $h(x) = \sqrt{x + 4}$ was found to be continuous on the closed interval $[-4, 4]$ and differentiable on the open interval $(-4, 4)$. The function is continuous for $x \geq -4$, and its derivative, $\frac{1}{2\sqrt{x + 4}}$, exists for $x > -4$. Thus, $h(x)$ satisfies the conditions of the MVT.

Therefore, based on our analysis, functions I and III satisfy the conditions of the Mean Value Theorem over the interval $[-4, 4]$. This leads us to the final conclusion:

The correct answer is D. I and III only.

This comprehensive analysis underscores the importance of carefully checking both continuity and differentiability when applying the Mean Value Theorem. Failing to verify these conditions can lead to incorrect conclusions and a misunderstanding of the theorem's implications.