91Ó°ÊÓ

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0} .\) You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section \(4.8) .\) Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0},\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} f^{\prime}(x) .\) $$ \begin{array}{l}{\text { Show that }} \\ {\qquad f(x)=\left\\{\begin{array}{ll}{x^{2}+x+1,} & {x \leq 1} \\ {3 x,} & {x>1}\end{array}\right.}\end{array} $$ $$ \begin{array}{l}{\text { is continuous at } x=1 . \text { Determine whether } f \text { is differen- }} \\ {\text { tiable at } x=1 . \text { If so, find the value of the derivative there. }} \\ {\text { Sketch the graph of } f}\end{array} $$

Step by step solution

01

Determine Continuity at x=1

First, check that the left-hand and right-hand limits at \(x=1\) are equal to the function's value at \(x=1\). For the left-hand limit where \(x \leq 1\), evaluate \(\lim_{x \to 1^-} (x^2 + x + 1)\). For the right-hand limit where \(x > 1\), evaluate \(\lim_{x \to 1^+} 3x\). Both limits must equal \(f(1)\).

02

Calculate Left-Hand Limit at x=1

Evaluate \(\lim_{x \to 1^-} (x^2 + x + 1) \). Plug \(x=1\) into the expression to get \(1^2 + 1 + 1 = 3\). Thus, the left-hand limit is \(3\).

03

Calculate Right-Hand Limit at x=1

Evaluate \(\lim_{x \to 1^+} 3x \). Plug \(x=1\) into the expression to get \(3 \times 1 = 3\). Thus, the right-hand limit is \(3\).

04

Check Continuity Condition

The left-hand limit, right-hand limit, and \(f(1)\) are all \(3\). Thus, \(f\) is continuous at \(x = 1\).

05

Calculate Derivatives for Each Piece

Find \(f'(x)\) for each piece. For \(x^2 + x + 1\), the derivative is \(2x + 1\) for \(x \leq 1\). For \(3x\), the derivative is \(3\) for \(x > 1\).

06

Evaluate the Limit of the Derivative as x Approaches 1

Calculate \(\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} (2x + 1) = 2 \times 1 + 1 = 3\), and \(\lim_{x \to 1^+} f'(x) = 3\). Both limits are \(3\).

07

Ensure Differentiability at x=1

Since the left-hand and right-hand limits of the derivative are equal and exist, \(f\) is differentiable at \(x = 1\), and \(f'(1) = 3\).

08

Sketch the Graph of the Function

Sketch the graph by plotting \(f(x) = x^2 + x + 1\) for \(x \leq 1\) and \(f(x) = 3x\) for \(x > 1\). Ensure continuity and notice the smooth transition with the derivative value at \(x = 1\) being consistent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Piecewise-defined functions

In calculus, piecewise-defined functions are functions composed of multiple sub-functions, each specified for different parts of the domain. These sub-functions are expressed with conditions, indicated by using different expressions for different ranges of the independent variable.

To illustrate, consider a function described as follows:

• For values of \( x \leq 1 \), the function is described by \( f(x) = x^2 + x + 1 \).
• For \( x > 1 \), the function becomes \( f(x) = 3x \).

Piecewise functions are useful when different rules are applied under distinct scenarios. They can model real-world situations with various changing conditions. Understanding how these functions behave across their domain is crucial, especially at points where the sub-functions change, as these are points of interest for questions regarding continuity and differentiability.

Continuity at a point

Continuity at a point in a function means that as you approach that point from either side, the function appropriately "connects" or behaves predictably without any jumps or breaks. In technical terms, a function \( f(x) \) is continuous at a point \( x_0 \) if the following conditions hold:

• The function is defined at \( x_0 \), meaning \( f(x_0) \) exists.
• The left-hand limit, \( \lim_{x \to x_0^-} f(x) \), and the right-hand limit, \( \lim_{x \to x_0^+} f(x) \), exist and are equal.
• Both limits equal \( f(x_0) \).

For the piecewise function example we're discussing, continuity at \( x = 1 \) is verified by showing that the left-hand limit (from the side with \( x^2 + x + 1 \)) and the right-hand limit (from \( 3x \)) are both equal to \( 3 \), and also equal to the actual function value \( f(1) = 3 \).

Ensuring these conditions are met confirms that both branches of the piecewise function meet smoothly at \( x = 1 \), indicating continuity.

Derivative calculation

Calculating the derivative of a function involves finding an expression that gives the rate at which the function's value changes. For a piecewise function, it's crucial to compute the derivative for each of the sub-functions separately.

For the given function, determine the derivatives as follows:

• For \( f(x) = x^2 + x + 1 \) where \( x \leq 1 \), the derivative is \( f'(x) = 2x + 1 \).
• For \( f(x) = 3x \) where \( x > 1 \), the derivative is simply \( f'(x) = 3 \).

Next, evaluate the behavior of these derivatives at the point of interest \( x = 1 \), where the sub-functions switch. It's crucial to analyze both the left and right limits of the derivative as \( x \) approaches 1:

• Calculating \( \lim_{x \to 1^-} (2x + 1) = 3 \).
• Similarly, \( \lim_{x \to 1^+} 3 = 3 \).

Since these limits are equal, this indicates that the derivative exists at \( x = 1 \), making the function differentiable at this point.

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental theorem in calculus that links the concepts of continuity and differentiability of functions. It states that for a continuous function on \([a, b]\) that is differentiable on \((a, b)\), there exists at least one point \( c \) in the open interval \((a, b)\) such that:\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]This theorem is often used to justify claims about a function's behavior within an interval.

In the context of finding differentiability at \( x = 1 \) for our piecewise function, the theorem assists indirectly by ensuring that, under specific conditions, having equal left-hand and right-hand derivatives implies the smooth, unbroken nature of the function at that point. Here, since both the left and right derivatives exist and are equal at \( x = 1 \), the function fulfills the differentiability criteria outlined by the horizontal and vertical considerations of the theorem.