Problem 6 - 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 \(x_{1}\). $$ f(x)= \begin{cases}x & \text { if } x \leq 0 \\ x^{2} & \text { if } x>0 \\\ x_{1}=0\end{cases} $$

Step-by-Step Solution

Answer

The function is continuous at \(x_1 = 0\) but not differentiable at \(x_1 = 0\).

Step 1: Draw the Graph

Plot the graph of the function. The function is defined as follows: For \(x \leq 0\), \[f(x) = x\]. This is a straight line through the origin with a slope of 1. For \(x > 0\), \[f(x) = x^2\]. This is a parabolic curve opening upwards starting from the origin.

Step 2: Determine Continuity at \(x_1 = 0\)

To check if \(f\) is continuous at \(x_1 = 0\), we need to ensure that \[\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^+}} f(x) = f(0) \]. First, calculate \[f(0)\]: \[f(0) = 0\]. Next, \[\lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^-}} x = 0\]. Finally, \[\lim_{{x \to 0^+}} f(x) = \lim_{{x \to 0^+}} x^2 = 0\]. Since all three values are equal, the function is continuous at \(x_1 = 0\).

Step 3: Find the Left-Hand Derivative \(f'_{-}(0)\)

Calculate the left-hand derivative at \(x_1 = 0\). The function for \(x \leq 0\) is \[f(x) = x\]. The left-hand derivative is \[f'_{-}(0) = \lim_{{x \to 0^-}} \frac{{f(x) - f(0)}}{{x - 0}} = \lim_{{x \to 0^-}} \frac{{x - 0}}{{x}} = 1\].

Step 4: Find the Right-Hand Derivative \(f'_{+}(0)\)

Calculate the right-hand derivative at \(x_1 = 0\). The function for \(x > 0\) is \[f(x) = x^2\]. The right-hand derivative is \[f'_{+}(0) = \lim_{{x \to 0^+}} \frac{{f(x) - f(0)}}{{x - 0}} = \lim_{{x \to 0^+}} \frac{{x^2 - 0}}{{x}} = \lim_{{x \to 0^+}} x = 0\].

Step 5: Determine Differentiability at \(x_1 = 0\)

To determine if \(f\) is differentiable at \(x_1 = 0\), the left-hand and right-hand derivatives must be equal. From Step 3, \[f'_{-}(0) = 1\]. From Step 4, \[f'_{+}(0) = 0\]. Since \[f'_{-}(0) \eq f'_{+}(0)\], \[f\] is not differentiable at \(x_1 = 0\).

Key Concepts

Graph Sketching

When sketching graphs for piecewise functions, it helps you understand their behavior better. For the given function, two parts need to be drawn:

• For \(x \leq 0\), \(f(x) = x\). This is a straight line passing through the origin with a slope of 1.
• For \(x > 0\), \(f(x) = x^2\). This part is a parabola opening upwards. It touches the origin but continues upwards as \(x\) increases.

Seeing both parts of the graph makes it easier to visualize the function's properties, such as continuity and differentiability.
The graph shows a transition at \(x = 0\), making it clear where you should check for these properties.

Limit Definition

The concept of limits is fundamental in understanding the behavior of functions at specific points. To check if \(f\) is continuous at \(x_1 = 0\), we use the limit definition. First, find \(f(0)\):
\[f(0) = 0\] Next, evaluate the left-hand limit as \(x\) approaches 0 from the left:
\( \lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^-}} x = 0 \). Then, find the right-hand limit as \(x\) approaches 0 from the right:
\( \lim_{{x \to 0^+}} f(x) = \lim_{{x \to 0^+}} x^2 = 0 \). Since these limits equal the value of the function at that point, it is continuous at \(x_1 = 0\). Continuity means no jumps or holes in the graph at the point.

Left-Hand Derivative

To find the left-hand derivative at \(x_1=0\), use the definition of the derivative approaching from the left. The function for \(x \leq 0\) is \(f(x) = x\).
Use the formula for the left-hand derivative:
\[f'_{-}(0) = \lim_{{x \to 0^-}} \frac{{f(x) - f(0)}}{{x-0}} = \lim_{{x \to 0^-}} \frac{{x-0}}{{x}} = 1\]
The left-hand derivative is 1. This result means that as \(x\) approaches 0 from the left side, the slope of the function is 1. Visually, this shows a line with a 45-degree angle to the \(x\)-axis on the left side of the origin.

Right-Hand Derivative

The right-hand derivative at \(x_1=0\) requires evaluating the function approaching from the right. For \(x > 0\), \(f(x)=x^2\).
Apply the formula for the right-hand derivative:
\[f'_{+}(0) = \lim_{{x \to 0^+}} \frac{{f(x) - f(0)}}{{x - 0}} = \lim_{{x \to 0^+}} \frac{{x^2 - 0}}{{x}} = \lim_{{x \to 0^+}} x = 0\] So, the right-hand derivative is 0. This result indicates that as \(x\) approaches 0 from the right side, the slope of the function tends to 0. This flat slope is characteristic of the bottom of the parabolic curve at the origin.

Differentiability

Finally, to determine differentiability at \(x_1=0\), compare the left-hand and right-hand derivatives. From the previous steps, we found:

• Left-hand derivative: \(f'_{-}(0) = 1\)
• Right-hand derivative: \(f'_{+}(0) = 0\)

For a function to be differentiable at a point, these derivatives must be equal. Since \(1 \eq 0\), the function is not differentiable at \(x_1=0\). This makes sense visually: the left side forms a straight line, while the right side forms a curve, creating a sharp change in direction at the origin. Therefore, the function has a corner at \(x_1=0\), which disrupts smoothness and thus differentiability.