Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

In each of the Problems 1-21, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of c; if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ g(x)=|x| ;[-2,2] $$

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply because \( g(x) = |x| \) is not differentiable at \( x = 0 \).

Step 1: Verify Continuity in the Interval

The function \( g(x) = |x| \) is continuous on the interval \([-2, 2]\) because absolute value functions are continuous everywhere.

Step 2: Verify Differentiability in the Interval

To apply the Mean Value Theorem, the function must be differentiable on the open interval \((-2, 2)\). The function \( g(x) = |x| \) is not differentiable at \( x = 0 \) because the derivative from the left is \(-1\) and from the right is \(1\). Therefore, \( g(x) \) is not differentiable at \( x = 0 \).

Step 3: Conclusion on MVT Applicability

Since the function \( g(x) = |x| \) is not differentiable at \( x=0 \), the Mean Value Theorem does not apply to this function on the interval \([-2, 2]\).

Step 4: Sketch the Graph of the Function

To sketch \( g(x) = |x| \), note that it forms a "V" shape with the vertex at the origin. At \(x = -2\), \(g(x) = 2\), and at \(x = 2\), \(g(x) = 2\). It is non-differentiable at \(x = 0\), causing the "V" shape, with lines at \(45^\circ\) angles from the x-axis.

Key Concepts

Continuity

Continuity of a function is an essential concept, particularly when discussing the Mean Value Theorem (MVT). A function is considered continuous on a closed interval if it produces a smooth, unbroken graph over that range.
For the function \( g(x) = |x| \), it is important to know that absolute value functions are continuous everywhere. This means \( g(x) \) is continuous over the entire interval \([-2, 2]\).
In simple terms, you won't find any jumps, breaks, or holes in the graph of \( |x| \) over this range. Continuity ensures that each point smoothly connects to the next.

• This smooth connection is a requirement for the Mean Value Theorem to potentially be applicable.
• For any given interval, confirming continuity is a great first step in analyzing a function.

Differentiability

Differentiability goes hand in hand with continuity, but it takes the concept a step further. For a function to be differentiable at a point, it must be smooth and not have any sharp turns or cusps at that point. The derivative, or the slope of the tangent line, must exist.
In our exercise, \( g(x) = |x| \) is not differentiable at \( x = 0 \).
Here's why:

• From the left of \( x = 0 \), the slope is \(-1\), and from the right, it is \(+1\).
• This mismatch causes a cusp, making the function non-differentiable at that point.
• Thus, while \( g(x) \) is continuous at \( x = 0 \), it fails the differentiability test, excluding the Mean Value Theorem from being applicable.

It's crucial to always check if the function is both continuous and differentiable over the specified open interval when considering the MVT.

Absolute Value Function

Absolute value functions, like \( g(x) = |x| \), are commonly encountered in calculus. Understanding their properties is key to solving problems like the one in this exercise.
Some important points about absolute value functions include:

• They are defined as \( |x| = x \) when \( x \geq 0 \) and \( |x| = -x \) when \( x < 0 \).
• This nature means they form a distinct "V" shape on graphs.
• They are continuous throughout their domain but note the sharp point at the vertex (such as at the origin), which signals non-differentiability at that exact location.

Recognizing these characteristics helps in graph reasoning and in understanding the limitations regarding the application of the Mean Value Theorem in certain intervals.

Graph Sketching

Graph sketching is a valuable tool for visualizing functions and understanding their behavior. When sketching \( g(x) = |x| \), we have clear indicators for its shape and important points.
The absolute value function has a characteristic "V" shape:

• The vertex of the "V" is located at the origin \( (0,0) \).
• The arms of the "V" extend symmetrically upwards at \( 45^\circ \) angles from the x-axis, indicating constant slopes on either side when differentiability permits.
• At \( x = -2 \) and \( x = 2 \), the function values are equal to \( 2 \), showcasing symmetry.

Sketching this graph aids in visualizing how the function behaves across the interval \([-2, 2]\), helping to identify where differentiability fails and providing context for why the Mean Value Theorem cannot apply here.