Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

In each of the Problems 1-21, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of c; if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ f(x)=x^{2}+x ;[-2,2] $$

Step-by-Step Solution

Answer

The Mean Value Theorem applies, and the value of c is 0.

Step 1: Verify Continuity

The Mean Value Theorem (MVT) requires the function to be continuous on the closed interval \([-2, 2]\). The given function is \(f(x) = x^2 + x\). This is a polynomial function, and polynomial functions are continuous everywhere. Therefore, \(f(x)\) is continuous on \([-2, 2]\).

Step 2: Verify Differentiability

Next, we need to verify that the function is differentiable on the open interval \(-2, 2\). Again, since \(f(x) = x^2 + x\) is a polynomial, it is differentiable everywhere. Thus, it is differentiable on \(-2, 2\).

Step 3: Apply the Mean Value Theorem

Since the function satisfies both conditions for the MVT, we can apply it. According to the theorem, there exists at least one point \(c\) in the interval \(-2, 2\) such that \( f'(c) = \frac{f(2) - f(-2)}{2 - (-2)}\).

Step 4: Calculate f(a) and f(b)

Calculate \(f(-2)\) and \(f(2)\) using the function \(f(x) = x^2 + x\).\[ f(-2) = (-2)^2 + (-2) = 4 - 2 = 2\]\[ f(2) = (2)^2 + (2) = 4 + 2 = 6\]

Step 5: Calculate the Slope of the Secant Line

Now, calculate the slope of the secant line joining the points \(-2, f(-2)\) and \(2, f(2)\):\[ \frac{f(2) - f(-2)}{2 - (-2)} = \frac{6 - 2}{2 + 2} = \frac{4}{4} = 1\]

Step 6: Find f'(x)

Find the derivative of the function \(f(x) = x^2 + x\). Using basic differentiation rules:\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(x) = 2x + 1\]

Step 7: Solve f'(c) = 1 for c

Set the derivative equal to the slope of the secant line and solve for \(c\):\[ 2c + 1 = 1 \ 2c = 0 \ c = 0 \]

Step 8: Conclusion

The Mean Value Theorem applies to \(f(x) = x^2 + x\) on \([-2, 2]\), and the value of \(c\) that satisfies the theorem is \(c = 0\). This value lies within the open interval \(-2, 2\).

Key Concepts

Polynomial Functions

When diving into the world of mathematics, polynomial functions often serve as essential building blocks. These functions are expressions involving variables raised to whole number powers, combined using addition, subtraction, and multiplication. An example of a simple polynomial function is given by the formula: \( f(x) = x^2 + x \).
Polynomial functions have several key characteristics that make them particularly important:

• They are defined for all real numbers, meaning they have values at every point on the number line.
• Their graphs are smooth and continuous lines or curves, without any breaks or abrupt changes.
• The degree of a polynomial, which is the highest power of the variable present, tells us a lot about its shape and the number of possible intersections with the x-axis.

Because we're dealing with the polynomial \( f(x) = x^2 + x \), we know it's quadratic, with a degree of 2. This implies its graph is a parabola, which opens upwards. Recognizing these traits of polynomial functions is crucial for applying calculus concepts effectively.

Differentiability

Differentiability is a core concept in calculus that tells us whether a function has a derivative at each point in its domain. When we say a function is differentiable, it simply means that the derivative exists and is finite everywhere on the specified interval. For polynomial functions, such as \( f(x) = x^2 + x \), differentiability is guaranteed because:

• The function is smooth, without sharp corners or cusps.
• It has no holes or interruptions in the graph.
• The limits that define the derivative exist at every point.

To find the derivative of \( f(x) = x^2 + x \), we use basic rules of differentiation:\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(x) = 2x + 1 \].
This derivative is valid for any real number, affirming the function is differentiable across its entire domain, including the open interval \(-2, 2\) for our exercise. Understanding differentiability helps us find slopes at specific points, which is critical in applying the Mean Value Theorem.

Continuity

Continuity is a property of functions that ensures there are no breaks, jumps, or holes in their graphs. A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. More generally, a function is continuous over an interval if it's continuous at every point within that interval.
Polynomial functions, like \( f(x) = x^2 + x \), are naturally continuous everywhere because of how they're constructed:

• They involve only basic operations (addition, subtraction, multiplication) with integer powers.
• There are no division by zero issues or abrupt discontinuities.

For the interval \([-2, 2]\), our function \( f(x) = x^2 + x \) is continuous, which is a crucial requirement for applying the Mean Value Theorem. Without continuity, the theorem wouldn't hold, as the behavior of the function's graph could be unpredictable. Ensuring continuity sets the stage for exploring how the function changes smoothly over the interval.

Calculus

Calculus is a branch of mathematics focusing on change and motion. It's indispensable in understanding how functions behave, especially looking at rates of change (differentiation) and accumulation of quantities (integration). One powerful tool in calculus is the Mean Value Theorem (MVT), which gives insight into the relationship between differentiability and continuity.The MVT states that for a continuous function \( f \) on a closed interval \([a, b]\) and differentiable on the open interval \( (a, b) \), there exists a point \( c \) in \( (a, b) \) such that:\[ f'(c) = \frac{f(b) - f(a)}{b-a} \].
This theorem essentially guarantees that at some point in the interval, the instantaneous rate of change is equal to the average rate of change over the interval.For our polynomial \( f(x) = x^2 + x \) over \([-2, 2]\), the theorem applies because:

• The function is continuous and differentiable on \([-2, 2]\).
• The calculated \( c = 0 \) satisfies the conditions, as demonstrated by the derivative equal to the slope of the secant line.

Understanding calculus concepts like the MVT helps us peek into the underlying patterns and behaviors of functions, revealing how they change and evolve over their domains.