Problem 5 - 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 \(x_{1}\). $$ \begin{gathered} f(x)= \begin{cases}-1 & \text { if } x<0 \\ x-1 & \text { if } x \geq 0\end{cases} \\ x_{1}=0 & \end{gathered} $$

Step-by-Step Solution

Answer

1. Sketch the graph. 2. \(f\) is continuous at \(x = 0\). 3. \(f'(0^-) = 0\), \(f_+'(0) = 1\). 4. \(f\) is not differentiable at \(0\).

Step 1: Sketch the graph of the function

Draw the graph for the following piecewise function: For values of \(x < 0\), \(f(x) = -1\), which is a horizontal line at \(y = -1\). For values of \(x \geq 0\), \(f(x) = x-1\), which is a line with slope 1 and y-intercept at \(y = -1\).

Step 2: Determine Continuity at \(x = 0\)

To determine if \(f\) is continuous at \(x = 0\), we need to check the left-hand limit, right-hand limit, and the value of the function at \(x = 0\). 1. Left-hand limit: \( \lim_{{x \to 0^-}} f(x) = -1\). 2. Right-hand limit: \( \lim_{{x \to 0^+}} f(x) = 0 - 1 = -1\). 3. Value of the function at \(x = 0\): \( f(0) = 0 - 1 = -1\). Since all three values are equal, \(f\) is continuous at \(x = 0\).

Step 3: Find the Derivatives \(f'(0)\) and \(f_+'(0)\)

To find \(f'(0)\) and \(f_+'(0)\), we need to check the left-hand derivative and the right-hand derivative: 1. Left-hand derivative: \( f'(x) = 0\) for \( x < 0\), so \( f'(0^-) = 0\). 2. Right-hand derivative: \( f'(x) = 1\) for \( x \geq 0\), so \( f_+'(0) = 1\).

Step 4: Determine Differentiability at \(x = 0\)

A function is differentiable at a point if the left-hand derivative equals the right-hand derivative at that point. Since \( f'(0^-) = 0\) and \( f_+'(0) = 1\), the derivatives are not equal. Therefore, \(f\) is not differentiable at \(x = 0\).

Key Concepts

graph sketching

To sketch the graph of a piecewise function, you need to draw each part separately according to its defined interval. For the given function:

• For values of x < 0, f(x) = -1. This indicates a horizontal line at y = -1 extending to the left of the origin.
• For values of x ≥ 0, f(x) = x - 1. This represents a straight line starting at the point (0, -1) and increasing with a slope of 1.

By combining these, you get a piecewise graph: a horizontal line at -1 for negative x values and a line sloping upward from (0, -1) for non-negative x values.

continuity

A function is continuous at a point if its left-hand limit, right-hand limit, and function value at that point are all equal. To check the continuity of the function at x = 0:

• The left-hand limit as x approaches 0 from the left is \(\text{lim}_{x \to 0^-} f(x) = -1\).
• The right-hand limit as x approaches 0 from the right is \(\text{lim}_{x \to 0^+} f(x) = 0 - 1 = -1\).
• The value of the function at x = 0 is f(0) = 0 - 1 = -1.

Since all these values are -1, the function is continuous at x = 0.

left-hand limit

To understand the left-hand limit, consider the values of the function as x approaches the point from the left. For our function:

• As x approaches 0 from the left, we use the part of the function f(x) = -1. The limit is \(\text{lim}_{x \to 0^-} f(x) = -1\).

This helps determine the behavior of the function from the left of the point.

right-hand limit

When examining the right-hand limit, evaluate the values of the function as x approaches the point from the right. For our function:

• As x approaches 0 from the right, we use the part of the function f(x) = x - 1. The limit is \(\text{lim}_{x \to 0^+} f(x) = 0 - 1 = -1\).

This shows the behavior of the function from the right side of the point.

differentiatability

A function is differentiable at a point if its left-hand derivative equals its right-hand derivative at that point. To check this for our function at x = 0:

• The left-hand derivative is obtained by differentiating f(x) = -1 for x < 0, which is 0. Thus, f'(0^-) = 0.
• The right-hand derivative is obtained by differentiating f(x) = x - 1 for x ≥ 0, which is 1. Thus, f_+'(0) = 1.
• Since f'(0^-) ≠ f_+'(0), the function is not differentiable at x = 0.

Differentiability requires the derivatives from both sides to match at the point of interest.