Problem 10 - 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 \(x_{1}\). $$ f(x)= \begin{cases}x^{2} & \text { if } x<-1 \\ -1-2 x & \text { if } x \geq-1 \\ & x_{1}=-1 & \end{cases} $$

Step-by-Step Solution

Answer

The function is continuous and differentiable at \(x = -1\) with \(f'(-1) = -2\).

Step 1 - Draw the Sketch of the Function

Consider the piecewise function: \[ f(x)= \begin{cases} x^{2} & \text { if } x<-1 \-1-2 x & \text { if } x \geq -1 \end{cases} \]To draw the graph, plot the parabola \(x^2\) for \(x < -1\), and the linear function \(-1 - 2x\) for \(x \geq -1\). Observe the point where the pieces meet at \(x = -1\).

Step 2 - Determine Continuity at \(x_1 = -1\)

A function is continuous at a point if \(\lim_{{x \to x_1^-}} f(x) = \lim_{{x \to x_1^+}} f(x) = f(x_1).\)For \(x = -1\), calculate: \[ \lim_{{x \to -1^-}} f(x) = (-1)^2 = 1, \lim_{{x \to -1^+}} f(x) = -1 - 2(-1) = 1, \]Given \(f(-1) = -1 - 2(-1) = 1\), these limits are equal and match the function value at \(x = -1\). Therefore, \(f\) is continuous at \(x=-1\).

Step 3 - Find \(f'(-1)\) and \(f_+'(-1)\)

To find the derivative from the left at \(x = -1\), compute:\[ f'_{-}(-1) = \lim_{{h \to 0^-}} \frac{f(-1+h) - f(-1)}{h} = \lim_{{h \to 0^-}} \frac{(-1+h)^2 - 1}{h} = \lim_{{h \to 0^-}} \frac{(1 - 2h + h^2) - 1}{h} = \lim_{{h \to 0^-}} \frac{-2h + h^2}{h} = \lim_{{h \to 0^-}} (-2 + h) = -2 \]To find the derivative from the right at \(x = -1\), compute:\[ f'_{+}(-1) = \lim_{{h \to 0^+}} \frac{f(-1+h) - f(-1)}{h} = \lim_{{h \to 0^+}} \frac{(-1 - 2(-1+h)) - 1}{h} = \lim_{{h \to 0^+}} \frac{1 - 2h - 1}{h} = \lim_{{h \to 0^+}} (-2) = -2 \]Both left-hand derivative and right-hand derivative are equal to \(-2\). Thus, \(f'(-1) = -2\).

Step 4 - Determine Differentiability at \(x_1 = -1\)

A function is differentiable at a point if the derivatives from the left and right at that point exist and are equal. Since \( f'_{-}(-1) = f'_{+}(-1) = -2 \), the function is differentiable at \( x = -1 \).

Key Concepts

Piecewise Functions

A piecewise function is a function that is defined by different expressions depending on the value of the input. In the given problem, the function is defined as:

f(x) = \begin{cases} x^{2} & \text{if } x<-1 \ -1-2x & \text{if } x \geq -1 \end{cases}

To understand and analyze such functions, sketch the graph for each piece separately within their respective intervals. For example:

• The first piece, \({x^2}\), is a parabola and applies for \(x < -1\).
• The second piece, \(-1 - 2x\), is a linear function and applies for \(x\geq -1\).

Observing where these pieces meet, at \(x = -1\), often provides insight into the behavior of the function. This blend of different function rules helps us understand the overall shape and properties of the function.

Continuity

Continuity at a specific point means there are no gaps, jumps, or breaks in the function at that point. A function \(f(x)\) is continuous at \(x = x_1\) if the following condition holds: \(\lim_{{x \to x_1^-}} f(x) = \lim_{{x \to x_1^+}} f(x) = f(x_1)\). Putting it simply, the left-hand and right-hand limits of the function should both equal the function's value at that point.

For the given exercise at \(x = -1\):

• Left-hand limit: \(\lim_{{x \to -1^-}} f(x) = (-1)^2 = 1\).
• Right-hand limit: \(\lim_{{x \to -1^+}} f(x) = -1 - 2(-1) = 1\).

These limits are equal to each other, and the actual value of the function at \(x = -1\) is also \(1\). Therefore, the function is continuous at \(x = -1\). Understanding continuity helps ensure that our function behaves predictably without any breaks at specified points.

Derivatives

The derivative of a function at a specific point measures the rate at which the function value changes. For a function \(f(x)\), the derivative at \(x = x_1\) can be expressed as: \(f'(x_1) = \lim_{{h \to 0}} \frac{f(x_1 + h) - f(x_1)}{h}\).

In a piecewise function like the one in the exercise, you need to consider derivatives from both left and right sides of the point. These are called left-hand and right-hand derivatives.

• Left-hand derivative at \(x = -1\): \(f'_{-}(-1) = \lim_{{h \to 0^-}} \frac{f(-1+h) - f(-1)}{h}\). After simplification, this equals \(-2\).
• Right-hand derivative at \(x = -1\): \(f'_{+}(-1) = \lim_{{h \to 0^+}} \frac{f(-1+h) - f(-1)}{h}\). After simplification, this also equals \(-2\).

Both derivatives are equal, so the overall derivative \(f'(-1) = -2\). Understanding how the function changes can help us predict its behavior at other nearby points.

Left-Hand and Right-Hand Limits

Left-hand and right-hand limits are fundamental for studying the behavior of functions around specific points, especially for piecewise functions. For a point \(x = x_1\), the left-hand limit (denoted as \(\lim_{{x \to x_1^-}} f(x)\)) considers values approaching \(x_1\) from the left, while the right-hand limit (denoted as \(\lim_{{x \to x_1^+}} f(x)\)) considers values approaching from the right.

To evaluate these limits for the given function at \(x=-1\):

• Left-hand limit: \(\lim_{{x \to -1^-}} f(x) = (-1)^2 = 1\).
• Right-hand limit: \(\lim_{{x \to -1^+}} f(x) = -1 - 2(-1) = 1\).

Since these limits are equal, both approaching \(1\), it helps confirm the continuity of the function at that point. Left-hand and right-hand limits are especially important in determining whether a function behaves smoothly across points where its rule changes. They play a critical role in ensuring that functions are well-defined and predictable across their entire domain.