Problem 186 - 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. $$ [T] \quad y=\frac{1}{\sqrt{x+1}} \text { over }[0,3] $$

Step-by-Step Solution

Answer

The value of \(c\) is approximately 1.0801.

Step 1: Understanding the Function and the Interval

We are given the function \( y = \frac{1}{\sqrt{x+1}} \) to analyze over the interval \([0, 3]\). This involves plotting the curve of the function and finding a secant line over this interval. We need to use the Mean Value Theorem to determine the value of \( c \) in this interval.

Step 2: Graphing the Function and Secant Line

To graph the function in your calculator, consider the interval \([0, 3]\). When \( x = 0 \), \( y = \frac{1}{\sqrt{0+1}} = 1 \), and when \( x = 3 \), \( y = \frac{1}{\sqrt{3+1}} = \frac{1}{2} \). We also plot the secant line that passes through these points: (0, 1) and (3, 0.5). The slope of this secant line can be found using \( m = \frac{0.5 - 1}{3 - 0} = -\frac{0.5}{3} = -\frac{1}{6} \).

Step 3: Applying the Mean Value Theorem

The Mean Value Theorem states that if a function is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists some \( c \) in \((a, b)\) such that \( f'(c) = \frac{f(b) - f(a)}{b - a} \). For our function, we have \( f'(x) = -\frac{1}{2}(x+1)^{-3/2} \).

Step 4: Finding the Derivative Equal to the Secant's Slope

We set the derivative equal to the slope of the secant line: \[ -\frac{1}{2}(x+1)^{-3/2} = -\frac{1}{6} \] Solving for \( x \), we have: \[ (x+1)^{-3/2} = \frac{1}{3} \] Taking both sides to the power of \(-2/3\), we derive: \[ x+1 = 3^{2/3} \] Solving gives: \[ x = 3^{2/3} - 1 \]

Step 5: Calculating and Approximating \(c\)

Now, to determine \( c \), compute \( 3^{2/3} \) using your calculator to get approximately \( 2.08008 \). Therefore, \( x = 2.08008 - 1 = 1.08008 \). Rounding this to four decimal places gives \( c \approx 1.0801 \).

Key Concepts

Function Graphing

Understanding how to graph a function is crucial in visualizing its behavior over a specific interval. The function in the exercise is given as \( y = \frac{1}{\sqrt{x + 1}} \). This function is continuous on the interval \([0, 3]\), meaning that as we plot the values, the graph won't "jump" or break. It smoothly transitions from one point to the next.

When graphing, start by determining key values. At \( x = 0 \), \( y = \frac{1}{\sqrt{0 + 1}} = 1 \), and at \( x = 3 \), \( y = \frac{1}{\sqrt{3 + 1}} = \frac{1}{2} \). These points help establish the curve's path on this interval. Plot these points and connect them smoothly to visualize the entire function.

A graphing calculator is a helpful tool since it can plot the function over this interval quickly, showing its behavior between and beyond these calculated points.

• Verify the continuity and differentiability of the function in the interval.
• Identify critical points to accurately describe the changes in the graph.
• Use a calculator to refine the details of the graph.

Secant Line

A secant line offers a straightforward way to view the average rate of change of a function over an interval. In our scenario, the secant line connects the points (0, 1) and (3, 0.5).

To determine this line, we calculate its slope, which is the change in \( y \) over the change in \( x \). Using the slope formula \( m = \frac{\Delta y}{\Delta x} \), we find:
\[ m = \frac{0.5 - 1}{3 - 0} = -\frac{0.5}{3} = -\frac{1}{6} \]
This negative slope shows that as we move along the interval from \( x = 0 \) to \( x = 3 \), the function's values decrease.

The secant line essentially represents the average rate at which the function's value drops over the interval.

• Identify the endpoints of the interval on the graph.
• Calculate the slope using these endpoints.
• Understand that the slope of the secant line equates to the average change over the interval.

Derivative Calculation

The derivative is pivotal for finding instantaneous rates of change and is a key player in the Mean Value Theorem. For this function, the derivative is:
\[ f'(x) = -\frac{1}{2}(x+1)^{-3/2} \]
This expression gives the slope of the tangent line to the curve at any point \( x \) on the interval.

According to the Mean Value Theorem, somewhere within the interval, the slope of the tangent (given by the derivative) equals the slope of the secant line. For our function's derivative to match the secant’s slope \(-\frac{1}{6}\), solve:
\[ -\frac{1}{2}(x+1)^{-3/2} = -\frac{1}{6} \]
By solving this equation, we find that:
\[ (x+1)^{-3/2} = \frac{1}{3} \]
This step is about algebraic manipulation to find \( c \), the exact point where the Mean Value Theorem holds true.

Thus, solving the equation yields the critical point \( c \) you'll calculate or estimate with a calculator. This point offers a unique property of the function interval.

• Understand the concept of the derivative as a tangent slope.
• Relate the derivative slope to the secant line slope through the Mean Value Theorem.
• Solve algebraic expressions accurately to isolate and determine \( c \).