Problem 4 - The Calculus with Analytic Geometry

Question

Do each of the following: (a) Draw a sketch of the graph of the function; (b) determine if \(f\) is continuous at \(x_{1} ;\) (c) find \(f^{\prime}-\left(x_{1}\right)\) and \(f_{+}^{\prime}\left(x_{1}\right)\) if they exist; (d) determine if \(f\) is differentiable at \(x_{1}\). $$ f(x)=1+|x+2| ; x_{1}=-2 $$

Step-by-Step Solution

Answer

The function is continuous at \( x_1 = -2 \) but not differentiable at \( x_1 = -2 \). The left-hand derivative is \(-1\), and the right-hand derivative is \(1\).

Step 1: Sketch the Graph of the Function

First, note that the function given is: \[ f(x)=1+|x+2| \]. This is an absolute value function shifted vertically by 1 unit. To sketch it, consider the vertex of the absolute value function, which occurs at the point \(x+2=0\) or \(x=-2\). At \(x=-2\), we have \[ f(-2) = 1+|-2+2| = 1+0 = 1 \]. The graph will have a 'V' shape with the vertex at \((-2, 1)\). For \(x > -2\), the function behaves as \( f(x) = 1 + (x+2) \). For \(x < -2\), the function behaves as \( f(x) = 1 - (x+2) \).

Step 2: Check Continuity at \(x_1 = -2\)

A function is continuous at a point if the limit as \(x\) approaches the point from both sides equals the function value at that point. For \(x_1 = -2\), check the left-hand limit: \( \text{lim}_{x \to -2^-} f(x) = \text{lim}_{x \to -2} (1 - (x+2)) = 1 \) and the right-hand limit: \( \text{lim}_{x \to -2^+} f(x) = \text{lim}_{x \to -2} (1 + (x+2)) = 1 \). Since both limits equal \( f(-2) = 1 \), the function is continuous at \( x_1 = -2 \).

Step 3: Find \(f^{'-(x_1)}\) and \(f^{'+}(x_1)\)

To find the derivatives from the left and right at \(x_1 = -2\), use the piecewise definitions: For \(x > -2\), \( f(x) = 1 + (x + 2) \rightarrow f'(x) = 1 \). For \(x < -2\), \( f(x) = 1 - (x + 2) \rightarrow f'(x) = -1 \). Thus, \( f'^-(-2) = -1 \) and \( f'^+(x_1)=1 \).

Step 4: Determine if \(f\) is Differentiable at \(x_1 = -2\)

A function is differentiable at a point if the left-hand and right-hand derivatives at that point are equal. Here, \( f'^-(-2) = -1 \) and \( f'^+(-2) = 1 \). Since the left-hand and right-hand derivatives are not equal, \(f\) is not differentiable at \( x_1 = -2 \).

Key Concepts

Graph Sketching

When graphing an absolute value function, note its distinctive 'V' shape. For the function given, \[ f(x)=1+|x+2| \], the vertex occurs where the inside of the absolute value is zero, i.e., at \( x = -2 \). At this point, \[ f(-2)=1+|-2+2| = 1 + 0 = 1. \] This vertex is at \((-2, 1)\). For \( x > -2 \), the function is \[ f(x) = 1 + (x + 2), \] a straight line with a slope of 1. For \( x < -2 \), the function is \[ f(x) = 1 - (x + 2), \] a straight line with a slope of -1. These points and slopes ensure the graph forms a 'V' shape.

Continuity

A function is continuous at a point if it meets three criteria:

• The function is defined at the point.
• The limit of the function exists as \( x \) approaches the point.
• The function value equals this limit.

For \( x_1 = -2 \), the function value is \[ f(-2) = 1. \]The left-hand limit, as \( x \) approaches -2 from the left, is \[ \text{lim}_{x \to -2^-} f(x) = \text{lim}_{x \to -2} (1 - (x+2)) = 1. \]The right-hand limit is \[ \text{lim}_{x \to -2^+} f(x) = \text{lim}_{x \to -2} (1 + (x+2)) = 1. \]Since the limits from both sides are equal to the function value, \( f \) is continuous at \( x = -2 \).

Left-Hand Derivative

To find the left-hand derivative at the point \( x_1 = -2 \), we use the part of the function defined for \( x < -2 \).For these values, \[ f(x) = 1 - (x + 2). \]The derivative of this expression is \[ f'(x) = -1. \]Therefore, the left-hand derivative at \( x = -2 \) is \[ f'^- (-2) = -1. \]This quantifies how the function behaves at exactly that side of the point \( x_1 \).

Right-Hand Derivative

Next, we calculate the right-hand derivative at the same point \( x_1 = -2 \).For \( x > -2 \), the function is \[ f(x) = 1 + (x + 2). \]The derivative of this part is \[ f'(x) = 1. \]Thus, the right-hand derivative at \( x = -2 \) is \[ f'^+ (-2) = 1. \]This tells us how the function behaves just past the point on that side.

Differentiability

A function is differentiable at a point if its left-hand and right-hand derivatives at that point are equal. This means there's a smooth transition with no corner at that point. For \( x_1 = -2 \), we previously found that

• \( f'^- (-2) = -1 \)
• \( f'^+ (-2) = 1 \)

Since these derivatives are not equal, \( f \) has a sharp corner at \( x = -2 \) and is not differentiable there. Continuity alone does not guarantee differentiability, as seen in this scenario.