Problem 17 - The Calculus with Analytic Geometry

Question

(a) Draw a sketch of the graph of the given function on the indicated interval; (b) test the three conditions (i), (ii), and (iii) of the hypothesis of Rolle's theorem and determine which conditions are satisfied and which, if any, are not satisfied; and (c) if the three conditions in part (b) are satisfied, determine a point at which there is a horizontal tangent line. $$ f(x)=\left\\{\begin{array}{ll} x+3 & \text { if } x \leq 2 \\ 7-x & \text { if } 2

Step-by-Step Solution

Answer

The function is continuous, but not fully differentiable at \( x = 2 \), so Rolle's theorem does not apply.

Step 1 - Understand the Function

The function, denoted as \( f(x) \), is defined as a piecewise function. For \( x \leq 2 \), the function is \( f(x) = x + 3 \), and for \( x > 2 \), the function is \( f(x) = 7 - x \). The interval under consideration is \([-3, 7]\).

Step 2 - Draw the Graph

Plot the piecewise function on the interval \([-3, 7]\). For \( x \leq 2 \), plot the line \( f(x) = x + 3 \) from \( x = -3 \) to \( x = 2 \). For \( x > 2 \), plot the line \( 7 - x \) from \( x = 2 \) to \( x = 7 \). Make sure the graph is continuous, but check for discontinuities or points of discontinuity.

Step 3 - Check Continuity at the Boundary

Evaluate the function's value at the point where the pieces meet, \( x = 2 \). For \( x = 2 \), \( f(2) = 2 + 3 = 5 \). Ensure that both pieces of the function match at this point. For \( 2 < x \), \( f(x) = 7 – x \rightarrow f(2^+) = 5 \). Thus, the function is continuous at \( x = 2 \).

Step 4 - Check the Conditions of Rolle's Theorem

To apply Rolle's Theorem, the function must be continuous on the closed interval \( [-3, 7] \), differentiable on \((-3, 7)\), and \( f(-3) = f(7) \). (i) Continuity: The function is continuous on \( [-3, 7] \) as shown in Step 3. (ii) Differentiability: The function is differentiable on \((-3, 7)\) except possibly at \( x = 2 \), where the derivative might not be defined. (iii) Same values at endpoints: Evaluate \( f(-3) = -3 + 3 = 0 \) and \( f(7) = 7 - 7 = 0 \). Both endpoints return 0, so this condition is satisfied.

Step 5 - Determine Points with Horizontal Tangent Lines

Since the function meets the conditions of Rolle's Theorem except for possible non-differentiability at \( x = 2 \), find where the derivative is 0 on the interval. For \( x eq 2 \), differentiate each piece separately: \( f'(x) = 1 \) for \( x eq 2 \leq 2 \) and \( f'(x) = -1 \) for \( x > 2 \). Neither derivative equals 0, indicating no such point with a horizontal tangent line exists in this interval.

Key Concepts

piecewise functions

A piecewise function is a function that is defined by different expressions for different intervals of the domain. In this exercise, the function is defined as:
\[ f(x)=\begin{cases} x + 3 & \text{if } x \leq 2 \ 7 - x & \text{if } x > 2 \end{cases} \]
This function changes its expression at the point \( x = 2 \), creating two distinct pieces. Such functions are particularly useful in modeling scenarios where a rule or relationship changes based on the input value.
When dealing with piecewise functions, it's important to:

• Clearly define the intervals for each piece.
• Ensure that the intervals cover all possible values in the domain without overlapping.
• Check for continuity and differentiability at the points where the function changes its expression.

graph sketching

Graph sketching involves plotting the given function on the specified interval. For our function:
1. For \( x \leq 2 \): The graph follows the line \( y = x + 3 \).
2. For \( x > 2 \): The graph follows the line \( y = 7 - x \).
To sketch the graph:

• Draw the line \( f(x) = x + 3 \) from \( x = -3 \) to \( x = 2 \).
• Ensure a smooth transition at \( x = 2 \) by checking the function's value at this point. For both pieces, \( f(2) = 5 \).
• Draw the line \( f(x) = 7 - x \) from \( x = 2 \) to \( x = 7 \).

It's important to pay attention to the behavior at \( x = 2 \), ensuring no gaps or jumps, which confirms continuity at that point.

differentiability

Differentiability of a function means that its derivative exists at each point in its domain, except possibly at points of piecewise definition changes.
For our function:
1. In the interval \( x \leq 2 \), the derivative is \( f'(x) = 1 \).
2. In the interval \( x > 2 \), the derivative is \( f'(x) = -1 \).
The piecewise nature leads us to check differentiability at the boundary \( x = 2 \). Here, the derivatives on either side differ:

For \( x \leq 2 \), \( f'(2) = 1 \).

For \( x > 2 \), \( f'(2) = -1 \).

This indicates a cusp or corner at \( x = 2 \), meaning the function is not differentiable at that point.

continuity

A function is continuous if there are no breaks, jumps, or holes in its graph over the interval.
Checking the continuity of our piecewise function involves ensuring the function's value is the same from both sides at \( x = 2 \):

• For \( x \leq 2 \), \( f(2) = 2 + 3 = 5 \).
• For \( x > 2 \), \( f(2) = 7 - 2 = 5 \).

Since both pieces agree at \( x = 2 \), the function is continuous at \( x = 2 \). Thus, the function is continuous over the entire interval \( [-3, 7] \). Continuity is a necessary condition for applying Rolle's Theorem, ensuring a seamless graph without disjointed segments.