Problem 188 - Calculus Volume 1

Question

Use a calculator to graph the function over the interval \([a, b]\) and graph the secant line from \(a\) to \(b .\) Use the calculator to estimate all values of \(c\) as guaranteed by the Mean Value Theorem. Then, find the exact value of \(c,\) if possible, or write the final equation and use a calculator to estimate to four digits. $$ [\mathrm{T}] \quad y=x+\frac{1}{x} \text { over }\left[\frac{1}{2}, 4\right] $$

Step-by-Step Solution

Answer

The value of \( c \) is \( \sqrt{2} \approx 1.414 \).

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem states that for a function \( f \) that is continuous over \( [a, b] \) and differentiable over \((a, b)\), there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). We first confirm that \( y = x + \frac{1}{x} \) meets these conditions.

Step 2: Calculate the Function Values

Calculate \( f(\frac{1}{2}) \) and \( f(4) \):- \( f(\frac{1}{2}) = \frac{1}{2} + \frac{1}{(\frac{1}{2})} = \frac{1}{2} + 2 = \frac{5}{2} \).- \( f(4) = 4 + \frac{1}{4} = 4.25 \).

Step 3: Find the Slope of the Secant Line

The slope of the secant line is found using the difference quotient:\[ \frac{f(4) - f(\frac{1}{2})}{4 - \frac{1}{2}} = \frac{4.25 - 2.5}{3.5} = \frac{1.75}{3.5} = 0.5 \].

Step 4: Differentiate the Function

Differentiate \( y = x + \frac{1}{x} \):\( \frac{dy}{dx} = 1 - \frac{1}{x^2} \).

Step 5: Set Derivative Equal to Secant Slope

Set the derivative equal to the slope of the secant line and solve for \( c \):\[ 1 - \frac{1}{c^2} = 0.5 \].

Step 6: Solve for \( c \)

Rearrange and solve for \( c \):\[ 1 - 0.5 = \frac{1}{c^2} \rightarrow 0.5 = \frac{1}{c^2} \rightarrow c^2 = 2 \rightarrow c = \sqrt{2} \approx 1.414 \].

Step 7: Validate the Solution

Since \( c = \sqrt{2} \) is in the interval \((\frac{1}{2}, 4)\), it is a valid solution that satisfies the Mean Value Theorem.

Key Concepts

Secant Line

The secant line is an important concept used in calculus to understand average rates of change within a specified interval. Imagine you are tracking the growth of a plant over time. The secant line would help you find its average growth rate between two points in time.To determine the slope of a secant line for a given function, you take the difference in the function's values at two points and divide that by the difference in the points themselves. Using the exercise example, we calculated the slope of the secant line for the function \( y = x + \frac{1}{x} \) over the interval \( \left[ \frac{1}{2}, 4 \right] \) as 0.5. This slope represents the average rate at which the function changes between these points.The Mean Value Theorem, which involves these secant lines, assures us that there is at least one point where the instantaneous rate of change (or the slope of the tangent line) is the same as this average rate of change.

Differentiation

Differentiation is a key tool in calculus, used to find the rate at which a function changes at any given point. It tells us the function's slope or how steep it is at exactly one spot.In our exercise, the function \( y = x + \frac{1}{x} \) was differentiated to get the derivative \( \frac{dy}{dx} = 1 - \frac{1}{x^2} \). This derivative represents the rate of change of the function's value as \( x \) alters.Differentiation allows you to zoom in on a specific section of a curve and understand its behavior precisely. By comparing this derivative to the secant line's slope, the Mean Value Theorem states there is at least one point \( c \) on the interval where the function's instantaneous rate of change matches the average rate of change. This concept is pivotal in understanding how functions behave, indicating peaks, valleys, and where curves change direction.

Continuous Function

A continuous function is a function without any breaks, holes, or gaps in its domain. Imagine drawing a line without lifting your pen from the paper—this illustrates continuity.The function \( y = x + \frac{1}{x} \) is continuous over the interval \( \left[ \frac{1}{2}, 4 \right] \). No sudden jumps or breaks occur within this range, fulfilling the necessary condition for the Mean Value Theorem to be applicable.Continuous functions are vital in calculus, as they ensure that within a given interval, each point transitions smoothly to the next. This property allows for accurate predictions about the function's behavior and confirms that calculus tools like the Mean Value Theorem can be applied effectively.

Differentiable Function

A differentiable function means that its derivative exists at each point within a given interval. In simpler terms, you can find a tangent line at every point of a differentiable function.For the Mean Value Theorem to hold, a function must be differentiable on the open interval it is studied. In our exercise, the function \( y = x + \frac{1}{x} \) was found to possess a derivative \( \frac{dy}{dx} = 1 - \frac{1}{x^2} \). This confirms its differentiability on the interval \( (\frac{1}{2}, 4) \), ensuring it is smooth with no sharp corners or discontinuities.Differentiable functions allow us to apply calculus techniques like finding maximums, minimums, and points of inflection, providing a detailed understanding of a function's characteristics.