Problem 18 - 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. $$ f(x)=\left\\{\begin{aligned} 3 x+6 & \text { if } x<1 \\ x-4 & \text { if } 1 \leq x \end{aligned}\right\\} ;[-2,4] $$

Step-by-Step Solution

Answer

Rolle's theorem does not apply since the function is not continuous at \(x = 1\).

Step 1: Understand the Given Function

The given function is a piecewise function with two parts: \(f(x) = 3x + 6\) if \(x < 1\) and \(f(x) = x - 4\) if \(x \geq 1\). The interval is \([-2,4]\).

Step 2: Sketch the Graph

First, sketch \(f(x) = 3x + 6\) for \(x < 1\). Next, sketch \(f(x) = x - 4\) for \(x \geq 1\). Ensure the graph is continuous by noting that both parts should meet at \(x = 1\). Calculate points for each segment as \(x\) approaches 1 from the left and right.

Step 3: Test Conditions of Rolle's Theorem

Verify that the function is continuous on \([-2, 4]\), differentiable on \((-2, 4)\), and that \(f(a) = f(b)\) where \(a = -2\) and \(b = 4\).

Step 4: Continuity Check

Check if \(f(x)\) is continuous on \([-2, 4]\). The only potential discontinuity is at \(x = 1\). Check \(\lim_{{x \to 1^-}} f(x) = 3(1) + 6 = 9\) and \(\f(x=1) = 1 - 4 = -3\). Since the values are different, the function is not continuous at \(x = 1\). This fails the continuity test.

Step 5: Differentiability Check (Skippable)

Since the function fails the continuity test, Rolle's theorem does not apply, and checking differentiability is not necessary.

Step 6: Conclusion

Since the function is not continuous at \(x = 1\), we do not need to look for a point where there is a horizontal tangent line as Rolle's theorem does not apply.

Key Concepts

Piecewise Function

A piecewise function is defined by different expressions for different intervals of its domain. In the given exercise, our function is defined as follows:
\[ f(x) = \begin{cases} 3x + 6 & \text{if } x < 1 \ x - 4 & \text{if } x \geq 1 \end{cases} \]
This means the function behaves differently depending on whether the value of \(x\) is less than 1 or greater than/equal to 1. When \(x < 1\), it follows the equation \(3x + 6\). When \(x \geq 1\), it follows \(x - 4\).
To thoroughly understand this type of function:

• Identify the different expressions and the intervals they apply to.
• Note the boundary points (in this case, \(x = 1\)).
• Sketch each part of the function within its specific interval.
• Ensure to consider how the parts connect at the boundary points.

Understanding how to sketch the graph and assess the function at boundary points is essential.

Continuity

Continuity is a key concept when dealing with piecewise functions. A function is continuous at a point if there is no interruption in the graph at that point. Mathematically, \(f\) is continuous at \(x = c\) if:

• \(f(c)\) is defined.
• \(\lim_{{x \to c}} f(x)\) exists.
• \(\lim_{{x \to c}} f(x) = f(c)\).

For our piecewise function, we need to check continuity at the point where the function formula changes (i.e., at \(x = 1\)). We compute the left-hand limit and the right-hand limit at \(x = 1\) and compare these with \(f(1)\):

• \(\lim_{{x \to 1^-}} (3x + 6) = 9\)
• \(\lim_{{x \to 1^+}} (x - 4) = -3\)
• \(f(1) = -3\)

Since \(\lim_{{x \to 1^-}} f(x) eq \lim_{{x \to 1^+}} f(x) \), the function is not continuous at \(x = 1\). Because of this, the function fails the continuity test for Rolle's theorem on the interval \([-2, 4]\). This failure means we cannot apply Rolle's theorem to this function on this interval.

Differentiability

Differentiability of a function at a point implies that its graph has a tangent there that is not vertical. If a function is differentiable at each point in an interval, it can be said to be differentiable on that interval.
For our function to apply Rolle's theorem, it must be differentiable on the open interval \((-2, 4)\). Considering the segments individually:

• For \(x < 1\), \(f(x) = 3x + 6\) is differentiable everywhere since it is a linear polynomial.
• For \(x \geq 1\), \(f(x) = x - 4\) is also differentiable everywhere since it is another linear polynomial.

However, differentiation requires continuity, which this function lacks at \(x = 1\). So, even if each segment is differentiable, the function as a whole is not continuous on \([-2, 4]\). Hence, checking differentiability is a moot point after continuity fails.

Horizontal Tangent Line

A horizontal tangent line occurs where the derivative of the function, \( f'(x) \), is zero. In other words, the slope of the tangent line at that point is flat. For Rolle's theorem, if the conditions are satisfied, there must be at least one point \(c\) in the interval \((a, b)\) where the derivative \( f'(c) = 0 \).
In typical problems involving Rolle's theorem:

• First, ensure the function is continuous on \([a, b]\).
• Next, verify the function is differentiable on \((a, b)\).
• Finally, confirm that \(f(a) = f(b)\).

For our function and interval \([-2, 4]\), the lack of continuity at \(x = 1\) disqualifies the function from satisfying the conditions of Rolle's theorem. Thus, we do not look for horizontal tangent lines here, as the theorem's requirements are unmet.
Continuity and differentiability are crucial for finding horizontal tangents using this method.