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Chapter 1: Problem 56

Let \(G\) be a function given by $$G(x)=\left\\{\begin{array}{ll}x^{3}, & \text { for } x \leq 1 \\ 3 x-2, & \text { for } x>1\end{array}\right.$$ a) Verify that \(G\) is continuous at \(x=1\). b) Is \(G\) differentiable at \(x=1 ?\) Why or why not?

Short Answer

Expert verified
The function \( G(x) \) is continuous and differentiable at \( x = 1 \).

Step by step solution

01

Understand the problem

The function \( G(x) \) is a piecewise function with different expressions based on the value of \( x \). We need to determine if \( G \) is continuous and differentiable at \( x = 1 \).
02

Define continuity at a point

For \( G(x) \) to be continuous at \( x = 1 \), the left-hand limit, right-hand limit, and the value of the function at \( x = 1 \) must all be equal.
03

Calculate left-hand limit as \( x \to 1^- \)

For \( x \leq 1 \), \( G(x) = x^3 \). Thus, the left-hand limit is \( \lim_{x \to 1^-} x^3 = 1^3 = 1 \).
04

Calculate right-hand limit as \( x \to 1^+ \)

For \( x > 1 \), \( G(x) = 3x - 2 \). Thus, the right-hand limit is \( \lim_{x \to 1^+} (3x - 2) = 3(1) - 2 = 1 \).
05

Evaluate \( G(1) \)

Since \( x \leq 1 \) gives us \( G(x) = x^3 \), we have \( G(1) = 1^3 = 1 \).
06

Verify continuity

Both limits and \( G(1) \) are equal to 1, so \( G(x) \) is continuous at \( x = 1 \).
07

Define differentiability at a point

For \( G(x) \) to be differentiable at \( x = 1 \), the derivative from the left and right of \( 1 \) must be equal.
08

Differentiate piecewise expressions

Differentiate \( x^3 \) to get \( 3x^2 \) for \( x \leq 1 \), and differentiate \( 3x - 2 \) to get \( 3 \) for \( x > 1 \).
09

Calculate left-hand derivative at \( x = 1 \)

The derivative as \( x \to 1^- \) is \( \lim_{x \to 1^-} 3x^2 = 3(1)^2 = 3 \).
10

Calculate right-hand derivative at \( x = 1 \)

The derivative as \( x \to 1^+ \) is \( 3 \).
11

Verify differentiability

The limit of the derivative from the left and right of \( x = 1 \) is equal, both being 3, so \( G(x) \) is differentiable at \( x = 1 \).

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Key Concepts

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

Continuity
To understand the continuity of a piecewise function like \(G(x)\) at a point, specifically at \(x = 1\), we need to verify three things:
  • The left-hand limit as \(x\) approaches \(1\): \(\lim_{x \to 1^-} G(x)\) must exist and be equal to the value of the function from the left side of 1, which, in this case, is given by \(x^3\).
  • The right-hand limit as \(x\) approaches \(1\): \(\lim_{x \to 1^+} G(x)\) must exist and be equal to the value of the function from the right side of 1, which is given by \(3x - 2\).
  • The actual value of the function at \(x = 1\): \(G(1)\) must be equal to both the left-hand and right-hand limits.
Here's how this checks out for the function \(G(x)\):
  • For the left-hand limit: \(\lim_{x \to 1^-} x^3 = 1^3 = 1\).
  • For the right-hand limit: \(\lim_{x \to 1^+} (3x - 2) = 3(1) - 2 = 1\).
  • The actual value of the function at \(x = 1\): \(G(1) = 1^3 = 1\).
Since all three values are 1, we can confidently say \(G(x)\) is continuous at \(x = 1\). Continuity ensures there is no sudden jump or break in the function at that point.
Differentiability
Differentiability of a function at a point means that the function can be smoothly drawn at that point, without any sharp corners or cusps. For a piecewise function like \(G(x)\), to be differentiable at \(x = 1\), both the left-hand and right-hand derivatives at \(x = 1\) should exist and be equal.Let's break down this process for \(G(x)\):
  • From \(x \le 1\): The derivative of \(x^3\) is \(3x^2\). Calculate the left-hand derivative at \(x = 1\): \(\lim_{x \to 1^-} 3x^2 = 3 \times 1^2 = 3\).
  • From \(x > 1\): The derivative of \(3x - 2\) is \(3\). Calculate the right-hand derivative at \(x = 1\): since the derivative is constant, it is \(3\).
Both derivatives give us a value of 3 at \(x = 1\), meaning the piecewise function \(G(x)\) is differentiable at \(x = 1\). The key takeaway is that no abrupt changes in slope exist at this point, which implies the function is smooth and continuous in its rate of change at \(x = 1\).
Limits
The concept of limits helps us understand the behavior of a function as it approaches a specific point. In the context of the given piecewise function \(G(x)\), understanding limits is critical for determining both continuity and differentiability.There are generally two types of limits to consider:
  • **Left-hand limit** \(\lim_{x \to c^-} f(x)\): This evaluates the behavior of \(f(x)\) as \(x\) approaches \(c\) from the left. For \(G(x)\) when \(x \le 1\), this is \(\lim_{x \to 1^-} x^3\), which is 1.
  • **Right-hand limit** \(\lim_{x \to c^+} f(x)\): This evaluates the behavior of \(f(x)\) as \(x\) approaches \(c\) from the right. For \(G(x)\) when \(x > 1\), this is \(\lim_{x \to 1^+} (3x - 2)\), which is also 1.
The crucial step is verifying that these two limits are indeed equal. If they are, the overall limit \(\lim_{x \to 1} G(x)\) exists, reinforcing the continuity and potential differentiability at the point \(x = 1\). Thus, limits serve as a foundational tool in checking the seamless progression and the smoothness of piecewise functions.

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Most popular questions from this chapter

Below are the steps in the simplification of the difference quotient for \(f(x)=\sqrt{x}\) (see Example 8). Provide a brief justification for each step. \(\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}\) a) \(=\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot\left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\) b) \(=\frac{x+h+\sqrt{x} \sqrt{x+h}-\sqrt{x} \sqrt{x+h}-x}{h(\sqrt{x+h}+\sqrt{x})}\) c) \(=\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\) d) \(=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}\) e) \(=\frac{1}{\sqrt{x+h}+\sqrt{x}}\)

Find the first through the fourth derivatives. Be sure to simplify each derivative before continuing. $$ f(x)=\frac{x+3}{x-2} $$

Find the interval(s) for which \(f^{\prime}(x)\) is positive. $$ f(x)=\frac{1}{3} x^{3}-x^{2}-3 x+5 $$

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\sqrt{7 x} $$

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 5}\left(\frac{x^{2}-25}{2 x-10}\right) $$
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