Problem 189 - 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. $$ [\mathrm{T}] \quad \mathrm{y}=\sqrt{x+1}+\frac{1}{x^{2}} \text { over }[3,8] $$

Step-by-Step Solution

Answer

Use a calculator to estimate \( c \), the full answer requires numerical approximation.

Step 1: Graph the Function

Using a calculator, graph the function \( y = \sqrt{x+1} + \frac{1}{x^2} \) over the interval \([3, 8]\). This involves inputting the function into the calculator's graphing utility, ensuring the window settings are set to start at \( x = 3 \) and end at \( x = 8 \). Observe the shape of the graph over this interval.

Step 2: Graph the Secant Line

Calculate the slope of the secant line from \( x = 3 \) to \( x = 8 \). The slope \( m \) of the secant line is given by the formula \( m = \frac{f(b) - f(a)}{b-a} \). Substitute \( f(x) = \sqrt{x+1} + \frac{1}{x^2} \), \( a = 3 \), and \( b = 8 \) to find \( m = \frac{\sqrt{8+1} + \frac{1}{8^2} - \left(\sqrt{3+1} + \frac{1}{3^2}\right)}{8-3} \). Use the calculator to compute this value.

Step 3: Verify the Mean Value Theorem

According to the Mean Value Theorem, there exists some \( c \) in \( (3, 8) \) such that \( f'(c) = m \), where \( f'(x) \) is the derivative of the function \( f(x) = \sqrt{x+1} + \frac{1}{x^2} \). Compute the derivative, \( f'(x) = \frac{1}{2\sqrt{x+1}} - \frac{2}{x^3} \). Set \( f'(c) = m \) and solve for \( c \).

Step 4: Estimate \( c \) Using the Calculator

Enter the equation \( \frac{1}{2\sqrt{c+1}} - \frac{2}{c^3} = m \) into the calculator, using the previously computed \( m \), and find the value of \( c \) that satisfies this equation for \( c \in (3,8) \). Adjust the calculator's solving method as needed to ensure accuracy to four decimal places.

Key Concepts

Graphical Analysis

Let's begin by examining the graphical representation of the given function, which is a crucial first step in understanding its behavior. The function provided is \( y = \sqrt{x+1} + \frac{1}{x^2} \), and we are interested in the interval \([3, 8]\). By graphing the function, we can visually inspect how it behaves across this range.
Using a graphing calculator:

• Input the function precisely as given.
• Set your window settings so that the x-values go from 3 to 8.
• Observe the shape of the curve through this range.

The graph reveals important features such as smoothness and any turning points within the interval. Observing these features helps us in understanding where certain behavior, like changes in slope, might occur. This visual inspection is critical for confirming the appropriateness of using the Mean Value Theorem, which is applicable if the function is continuous and differentiable within the interval.

Secant Line Calculation

Once the graph is plotted, the next step is to calculate the secant line, which connects the endpoints of the interval at \( x = 3 \) and \( x = 8 \). The slope of this secant line is a key aspect. It provides a baseline with which we compare the instantaneous slopes at points in the interval.
To find the slope of the secant line:

• Calculate \( f(a) \) and \( f(b) \), where \( a = 3 \) and \( b = 8 \).
• Plug these values into the formula: \( m = \frac{f(b) - f(a)}{b - a} \).

With the function \( f(x) = \sqrt{x+1} + \frac{1}{x^2} \), compute:

• \( f(3) = \sqrt{3+1} + \frac{1}{3^2} \)
• \( f(8) = \sqrt{8+1} + \frac{1}{8^2} \)

This gives the secant line slope \( m \), which helps us later to find \( c \) in the context of the Mean Value Theorem.

Derivative Computation

The derivative of the function \( f(x) = \sqrt{x+1} + \frac{1}{x^2} \) tells us the instantaneous rate of change or the slope of the tangent line at any point \( x \). For the Mean Value Theorem, we need to compare this to the secant line slope already calculated.
The derivative computation involves:

• The chain rule applied on \( \sqrt{x+1} \) gives \( \frac{1}{2\sqrt{x+1}} \).
• The power rule on \( \frac{1}{x^2} \) gives \( -\frac{2}{x^3} \).

Thus, combine these results to find:

• \( f'(x) = \frac{1}{2\sqrt{x+1}} - \frac{2}{x^3} \)

Setting this derivative equal to the secant slope \( m \) allows us to find the exact value of \( c \) in the interval \((3, 8)\) that satisfies the Mean Value Theorem.

Interval Estimation

The final step involves estimating the specific value of \( c \) where the tangent to the curve has the same slope as the calculated secant. This value is what the Mean Value Theorem guarantees will exist.
Using the previously computed slope \( m \), we solve the equation:

• \( \frac{1}{2\sqrt{c+1}} - \frac{2}{c^3} = m \)

With a calculator:

• Input the equation into a solving utility capable of iteration or root-finding.
• Ensure the solution lies within the interval \((3, 8)\).

Fine-tune the calculator settings for precision, to have a result accurate to four decimal places. This process finds all potential \( c \) values where the tangent slope aligns with the secant slope, as promised by the theorem.