Problem 8 - 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 \(x_{1}\). $$ \begin{gathered} f(x)= \begin{cases}x^{2}-4 & \text { if } x<2 \\ \sqrt{x-2} & \text { if } x \geq 2\end{cases} \\ x_{1}=2 \end{gathered} $$

Step-by-Step Solution

Answer

The function is continuous but not differentiable at \( x_1 = 2 \).

Step 1: Sketch the Graph

To sketch the graph of the function, draw two separate pieces. For the part of the function where \( x < 2 \), plot \( f(x) = x^2 - 4 \). For the part where \( x \geq 2 \), plot \( f(x) = \sqrt{x - 2} \). Ensure both parts meet at \( x = 2 \).

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

Check if the left-hand limit (LHL), right-hand limit (RHL), and the function's value at \( x_1 \) are equal. Calculate \( \lim_{{x \to 2^-}} f(x) \), \( \lim_{{x \to 2^+}} f(x) \), and \( f(2) \). If they are all equal, the function is continuous at \( x_1 \).

Step 3: Find \( f'_{-}(x_1) \)

Find the left-hand derivative at \( x_1 = 2 \). Use the definition of the derivative for the piece where \( x < 2 \). Calculate \( f'_{-}(2) = \lim_{{h \to 0^-}} \frac{f(2+h) - f(2)}{h} \).

Step 4: Find \( f'_{+}(x_1) \)

Find the right-hand derivative at \( x_1 = 2 \). Use the definition of the derivative for the piece where \( x \geq 2 \). Calculate \( f'_{+}(2) = \lim_{{h \to 0^+}} \frac{f(2+h) - f(2)}{h} \).

Step 5: Determine Differentiability at \( x_1 = 2 \)

Compare \( f'_{-}(2) \) and \( f'_{+}(2) \). If their values are equal, then the function is differentiable at \( x_1 = 2 \). Otherwise, it is not differentiable at that point.

Key Concepts

Continuity at a Point

A function's continuity at a given point means that the function's value at that point matches the expected value based on the surrounding points. To establish continuity at a specific point, say \(x_1\), three conditions must be met:

• The left-hand limit (LHL) as \(x\) approaches \(x_1\) must exist: \(\lim_{{x \to x_1^-}}f(x)\)
• The right-hand limit (RHL) as \(x\) approaches \(x_1\) must also exist: \(\lim_{{x \to x_1^+}}f(x)\)
• The function's value at the point should be equal to both limits: \(f(x_1)\)

For our example function at \(x_1 = 2\), this involves checking if the limits from the left-hand side and right-hand side of 2 are equal to the function's value at that point. Explicitly, we need \(\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) = f(2)\).

Left-Hand Limit

The left-hand limit (LHL) arises when we approach a specific point from the left side (i.e., smaller values) on the number line. In mathematical notation, we write this as \(\lim_{{x \to x_1^-}} f(x)\). For the provided piecewise function, to find the LHL as \(x\) approaches 2, we evaluate the segment where \(x < 2\). Here, \(f(x) = x^2 - 4\). So, we compute:
\[\lim_{{x \to 2^-}} (x^2 - 4)\]
Substitute \(x\) near 2 and realize it simplifies to:
\[\lim_{{x \to 2^-}} (x^2 - 4) = 2^2 - 4 = 4 - 4 = 0\]
Thus, the left-hand limit is 0.

Right-Hand Limit

Similarly, the right-hand limit (RHL) comes into play as we approach a point from the right side (larger values). This is represented by \(\lim_{{x \to x_1^+}} f(x)\). For our specific function where \(x \geq 2\), the relevant part is \(f(x) = \sqrt{x - 2}\). Now calculate:
\[\lim_{{x \to 2^+}} \sqrt{x - 2}\]
Substituting values close to 2 gives:
\[\lim_{{x \to 2^+}} \sqrt{x - 2} = \sqrt{2 - 2} = \sqrt{0} = 0\]
Therefore, the right-hand limit is also 0.

Left-Hand Derivative

The left-hand derivative at a point approaches it from the left side, utilizing the definition of the derivative. For \(x_1 = 2\), we use the function's piece where \(x < 2\). The derivative formula is:
\[f'_{-}(2) = \lim_{{h \to 0^-}} \frac{f(2+h) - f(2)}{h}\]
Substituting here, the expression becomes:
\[f'_{-}(2) = \lim_{{h \to 0^-}} \frac{(2+h)^2 - 4 - (2^2 - 4)}{h} = \lim_{{h \to 0^-}} \frac{4 + 4h + h^2 - 8 + 4}{h} = \lim_{{h \to 0^-}} \frac{4h + h^2}{h}\]
This simplifies to:
\[f'_{-}(2) = \lim_{{h \to 0^-}} (4 + h) = 4\]
Hence, the left-hand derivative is 4.

Right-Hand Derivative

The right-hand derivative at a point considers the function's behaviour as it nears the point from the right side. For \(x_1 = 2\), we use the segment \(x \geq 2\). The derivative definition is:
\[f'_{+}(2) = \lim_{{h \to 0^+}} \frac{f(2+h) - f(2)}{h}\]
Given \(f(x) = \sqrt{x - 2}\), compute:
\[f'_{+}(2) = \lim_{{h \to 0^+}} \frac{\sqrt{2 + h - 2} - \sqrt{2 - 2}}{h} = \lim_{{h \to 0^+}} \frac{\sqrt{h}}{h}\]
This expression evolves into:
\[f'_{+}(2) = \lim_{{h \to 0^+}} \frac{1}{\sqrt{h}}\]
As \(h\) approaches 0, \(\frac{1}{\sqrt{h}}\) becomes infinite. Thus, the right-hand derivative does not exist and is infinity.

Differentiability

A function is differentiable at a point if the derivative exists and is the same from both sides of that point. For differentiability at \(x_1 = 2\), compare the left-hand and right-hand derivatives:

• Left-hand derivative: \(f'_{-}(2)\) is 4.
• Right-hand derivative: \(f'_{+}(2)\) is infinite.

Because these derivatives are not equal, the function \(f\) is not differentiable at \(x_1 = 2\). Differentiability is contingent upon the function being continuous and the derivatives converging to the same value from either side, which isn't the case here.