Problem 21 - 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. $$ f(x)=\left\\{\begin{array}{ll} x^{2}-4 & \text { if } x<1 \\ 5 x-8 & \text { if } 1 \leq x \end{array}\right\\} ;\left[-2, \frac{8}{5}\right] $$

Step-by-Step Solution

Answer

Function is not continuous at x=1. Further steps for tangent can't be determined.

Step 1: Understand the Function Definition

The function is defined in two parts: For \( x < 1 \), \( f(x) = x^2 - 4 \). For \( 1 \leq x \), \( f(x) = 5x - 8 \).

Step 2: Sketch the Graph

Plot the pieces of the function within the given interval \([-2, \frac{8}{5}]\). - For \( x < 1 \), use the equation \( f(x) = x^2 - 4 \). - For \( 1 \leq x \), use the equation \( f(x) = 5x - 8 \). Connect these points smoothly to sketch the graph.

Step 3: Check Condition (i) of Rolle's Theorem

Condition (i) requires the function to be continuous on \([-2, \frac{8}{5}]\). Check if there are any discontinuities in given intervals. Check if the transition point at \( x = 1 \) is continuous.

Step 4: Check Condition (ii) of Rolle's Theorem

Condition (ii) requires the function to be differentiable on \((-2, \frac{8}{5})\). Check the differentiability at each point except the endpoint values. In particular, ensure the function is differentiable at \( x = 1 \).

Step 5: Check Condition (iii) of Rolle's Theorem

Condition (iii) requires \( f(a) = f(b) \) for the endpoints of the interval. Find \( f(-2) \) and \( f(\frac{8}{5}) \) and verify if they are equal.

Step 6: Find Horizontal Tangent Line

If all conditions are met, find \( c \) in \((a, b)\) such that \( f'(c) = 0 \). This derivative should be zero for an existing horizontal tangent line.

Key Concepts

Continuity

Continuity is one of the fundamental concepts required to apply Rolle's Theorem. A function is continuous on a closed interval \([a, b]\) if there are no breaks, jumps, or holes in its graph within that interval. To check if the given function is continuous:
\begin{array}{ll} x^{2}-4 & \text { if } x<1 \ 5 x-8 & \text { if } 1 \ \leqx \ \end{array}\

• The function is defined for two different expressions based on the value of \x.
• For \x < 1, the functional form is \f(x) = x^2 - 4.
• For \x \geq 1, it is \f(x) = 5x - 8.

This means we need to ensure both pieces meet seamlessly at \x = 1.
To check, calculate the limit of \f(x) from both sides as \x approaches 1.

• From the left: \lim_{{x \to 1^-}} f(x) = (1)^2 - 4 = -3
• From the right: \lim_{{x \to 1^+}} f(x) = 5(1) - 8 = -3.

Since \f(1^-)= f(1^+), the function is continuous at \x=1.
Therefore, \f(x) is continuous on the interval \([-2, \frac{8}{5}]\).

Differentiability

For a function to be differentiable on an open interval \(a, b\), it must have a defined and finite derivative at every point within that interval. Differentiability implies continuity, but the converse is not necessarily true. To investigate the function's differentiability on the interval \( -2, \frac{8}{5} \):

• For \x < 1: \f(x) = x^2 - 4. The derivative is \f' (x) = 2x.
• For \x \ \geq 1: \f(x) = 5x - 8. The derivative is \f' (x) = 5.

We now check the derivative's behavior at \x = 1, where the function pieces meet:

• From the left: \lim_{{x \to 1^-}} f'(x) = 2(1) = 2.
• From the right: \lim_{{x \to 1^+}} f' (x)= 5.

As \f'(1^-) ≠ f'(1^+), the function is not differentiable at \x = 1.
Consequently, the function is not differentiable over the entire interval, thus failing condition (ii) of Rolle's Theorem.

Horizontal Tangent

A horizontal tangent line occurs where the slope of the tangent line to the graph is zero. Mathematically, this condition is expressed as \f'(x) = 0. In the context of Rolle's Theorem, if all conditions (continuity, differentiability, and \f(a) = f(b)) are satisfied, then there exists a point \c in \(a, b\) such that \f'(c) = 0, indicating a horizontal tangent.
From our step-by-step solution, since the function is not differentiable at \x = 1, one of the primary conditions for Rolle's Theorem is not met. However, even without checking Rolle's Theorem, we might still find points where \f'(x) = 0 by examining within the domains of the pieces:

• For \x < 1, set \f'(x) = 0. Since \f'(x) = 2x, this gives \2x = 0 \Rightarrow x = 0.
• For \x \ \geq 1, set \f'(x) = 0. Since \f'(x) = 5, it never equals 0.

Thus, the only point satisfying \f'(c) = 0 is \x = 0, from the domain where \x<1. This suggests a horizontal tangent line at \x = 0 in the interval. Although Rolle’s Theorem is not applicable here, we identified where the slope is zero within the function.