Problem 19 - 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. $$ f(x)=\frac{x^{2}-x-12}{x-3} ;[-3,4] $$

Step-by-Step Solution

Answer

Rolle's theorem does not apply because the function is not continuous at \( x = 3 \).

Step 1: Simplify the Function

First, simplify the given function. Notice that we have a rational function: \( f(x) = \frac{x^2 - x - 12}{x - 3} \) Factor the numerator: \( x^2 - x - 12 = (x - 4)(x + 3) \), Therefore, \( f(x) = \frac{(x - 4)(x + 3)}{ x - 3} \). Simplify the function: \( f(x) = x + 3 \) for all \( x eq 3 \).

Step 2: Draw the Graph

Since the simplified function is linear for all values of \( x \) except at \( x = 3 \) (where it has a vertical asymptote), the graph consists of a straight line with a hole at \( x = 3 \). Plot \( f(x) = x + 3 \) over the interval [-3, 4] and include the vertical asymptote at \( x = 3 \).

Step 3: Verify Rolle's Theorem Conditions

Rolle's theorem has three conditions: (i) The function must be continuous on the closed interval \( [a, b] \).(ii) The function must be differentiable on the open interval \( (a, b] \).(iii) The function values at the endpoints \( f(a) \) and \( f(b) \) must be equal.Check condition (i): The function \( f(x) \) is not continuous at \( x = 3 \). Condition (i) is not satisfied.

Step 4: Condition Failure Implication

Since the function is not continuous over the entire interval, Rolle's theorem is not applicable.

Key Concepts

Continuity

Continuity is a fundamental concept in calculus. A function is continuous if there are no breaks, jumps, or holes in its graph. Formally, a function \(f(x)\) is continuous at a point \(x = c\) if:

• \(f(c)\) is defined.
• The limit of \(f(x)\) as \(x\) approaches \(c\) exists.
• The limit of \(f(x)\) as \(x\) approaches \(c\) is equal to \(f(c)\).

When checking the continuity of \(f(x) = \frac{x^2 - x - 12}{x - 3}\), after simplification to \(f(x) = x + 3\) except at \(x = 3\), we see it has a discontinuity (hole) at \(x = 3\). Therefore, \(f(x)\) is not continuous on the interval \([-3, 4]\) because of this point.

Differentiability

Differentiability means the function can be differentiated, or has a defined derivative, at each point in its domain. A function is differentiable at a point if it is smooth and has no corners or cusps at that point. For Rolle's theorem, the function must be differentiable on the open interval \((a, b)\). Since \(f(x)\) simplifies to a linear function \(x + 3\) for any \(x e 3\), it is well-behaved and differentiable everywhere except at the point of discontinuity (\(x = 3\)). Because \(f(x)\) is not continuous on the entire interval, differentiability cannot be fully considered relevant nor applied.

Interval Analysis

When analyzing intervals in calculus, particularly for Rolle's theorem, the function must be examined over a closed interval \([a, b]\) for continuity and over an open interval \((a, b)\) for differentiability. Here, we consider the interval \([-3, 4]\).
First, we find the function \(f(x) = \frac{x^2 - x - 12}{x - 3}\) after it is simplified. It turns into \(f(x) = x + 3\) except at \(x = 3\), where it has a hole.
Hence, when analyzing this interval, we recognize \(f(x)\) is not continuous at \(x = 3\). Because it doesn't meet the first criterion of Rolle's theorem (i.e., continuity over the entire closed interval), further interval analysis confirms that Rolle's theorem cannot be applied.

Graph Sketching

Graph sketching helps visualize the properties of functions. For the function \(f(x) = \frac{x^2 - x - 12}{x - 3}\), simplification to \(f(x) = x + 3\) reveals a linear function with a hole. This is due to the undefined nature at \(x = 3\).
To sketch the graph:

• Plot \(f(x) = x + 3\) as a linear function over the interval \([-3, 4]\).
• Identify that at \(x = 3\), there is a vertical asymptote (hole) since \(f(3)\) is undefined.
• Draw the line continuously but leave an open circle at \(x = 3\).

The graph visually highlights the discontinuity, helping students better understand why Rolle's theorem is not applicable here.