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Chapter 1: Problem 29

Find a value of the constant \(k,\) if possible, that will make the function continuous everywhere. $$ \text { (a) } f(x)=\left\\{\begin{array}{ll}{7 x-2,} & {x \leq 1} \\ {k x^{2},} & {x>1}\end{array}\right. $$ $$ \text { (b) } f(x)=\left\\{\begin{array}{ll}{k x^{2},} & {x \leq 2} \\ {2 x+k,} & {x>2}\end{array}\right. $$

Short Answer

Expert verified
(a) k = 5; (b) k = 4/3.

Step by step solution

01

Understand Continuity Criteria

A function is continuous at a point if the left and right limits at the point exist, are equal to each other, and are equal to the function's value at that point.
02

Evaluate Continuity for Part (a) at x = 1

For the function to be continuous at \( x = 1 \), ensure \( \lim_{{x \to 1^-}} f(x) = \lim_{{x \to 1^+}} f(x) = f(1) \). Calculate these limits:- \( \lim_{{x \to 1^-}} (7x - 2) = 5 \), because plugging \( x = 1 \) into \( 7x - 2 \) gives \( 5 \).- \( \lim_{{x \to 1^+}} (kx^2) = k \), since \( x^2 = 1 \) at \( x = 1 \).- To be continuous, set \( 5 = k \).
03

Calculate Continuity for Part (b) at x = 2

Similarly, check that \( \lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) = f(2) \).- \( \lim_{{x \to 2^-}} (kx^2) = 4k \), because \( (kx^2) \) at \( x = 2 \) becomes \( 4k \).- \( \lim_{{x \to 2^+}} (2x + k) = 4 + k \), plugging \( x = 2 \) into both expressions.- Setting \( 4k = 4 + k \), solve for \( k \).
04

Solve the Equation for k in Part (b)

Solve the equation derived from step 3: \( 4k = 4 + k \).- Rearrange to get \( 4k - k = 4 \).- Simplify to \( 3k = 4 \).- Divide both sides by 3, \( k = \frac{4}{3} \).
05

Final Result

For part (a), the continuous function needs \( k = 5 \). For part (b), \( k = \frac{4}{3} \) makes the function continuous.

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Key Concepts

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

Limits
Limits are a fundamental concept in calculus, especially when dealing with continuity. A limit is essentially the value that a function approaches as the input gets closer to a particular point. To determine continuity, it is crucial to check the limits from both the left side and the right side of the function's point of interest.
In the context of this exercise, understanding limits means:
  • Calculating the left-hand limit, which is what the function approaches as it comes from the lower side of the input value.
  • Calculating the right-hand limit, which shows the function's behavior as the input approaches from the upper side.
  • Ensuring these two limits are equal for a function to be continuous at that specific point.
In part (a), we find that as we approach 1 from the left with the expression \(7x - 2\), the limit is 5. From the right, with the expression \(kx^2\), it should equal the same value, determining the value of \(k\) required for continuity.
Piecewise Functions
Piecewise functions are defined by different expressions depending on the input value's range. Each "piece" of the function has its own definition and specific interval in which it applies. This makes evaluating piecewise functions slightly more complex because you need to consider each expression separately.
For these types of functions, you should:
  • Identify the points at which the definition of the function changes, often called the breakpoints.
  • Examine each piece of the function at the breakpoint, as both pieces need to meet seamlessly for the function to remain continuous.
  • Since different rules apply to different parts of the domain, ensure you work out the expressions individually.
In the problem above, notice the function changes its expression around \(x = 1\) and \(x = 2\) for parts (a) and (b), respectively.
Continuous Functions
A function is continuous at a point if there is no interruption, jump, or gap in the graph of the function at that point. The basic requirement is that the value of the function at a given point equals the value that the function is approaching from either side of that point. A continuous function will have no sudden jumps or breaks when graphed.
To ensure continuity, remember:
  • The left-hand limit must equal the right-hand limit at the point in question.
  • The value of the function at the point should be both finite and coincide with these limits.
  • If these conditions are met, the function remains smooth and unbroken over its domain.
In this exercise, continuity is tested by equating the left and right limits of each piecewise segment of the function at the transition points, leading to the determination of \(k\) for continuous behavior.
Calculus Problem Solving
Calculus problem solving often involves multiple steps and requires a thorough understanding of the principles involved, like limits and continuity. In these cases:
  • Break the problem into manageable parts.
  • Evaluate each condition that must hold true, such as setting equivalent limits for contiguous segments of a piecewise function.
  • Calculate any necessary constants like \(k\) to satisfy these continuity conditions.
In this specific problem, notice how setting the left and right limits equal is a direct application of calculus concepts to solve for the unknown \(k\), showcasing how fundamental calculus principles are used in practice.

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Most popular questions from this chapter

$$ \text { Find } \lim _{x \rightarrow \pi / 4} \frac{\tan x-1}{x-\pi / 4} $$

In an attempt to verify that \(\lim _{x \rightarrow 0}(\sin x) / x=1,\) a student constructs the accompanying table. (a) What mistake did the student make? (b) What is the exact value of the limit illustrated by this table? $$ \begin{array}{|c|c|c|c|c|}\hline x & {-0.01} & {-0.001} & {0.001} & {0.01} \\\ \hline \operatorname{sinx/x} & {0.017453} & {0.017453} & {0.017453} & {0.017453} \\ \hline\end{array} $$

In Example 5 we used the Squeezing Theorem to prove that $$ \lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=0 $$ Why couldn鈥檛 we have obtained the same result by writing $$ \begin{aligned} \lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right) &=\lim _{x \rightarrow 0} x \cdot \lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right) \\ &=0 \cdot \lim _{x \rightarrow 0} \sin \left(\frac{1}{x}\right)=0 ? \end{aligned} $$

Find the values of \(x\) (if any) at which \(f\) is not continuous, and determine whether each such value is a removable discontinuity. $$ \begin{array}{ll}{\text { (a) } f(x)=\frac{|x|}{x}} & {\text { (b) } f(x)=\frac{x^{2}+3 x}{x+3}} \\ {\text { (c) } f(x)=\frac{x-2}{|x|-2}}\end{array} $$

Find the limits. $$ \lim _{h \rightarrow 0} \frac{\sin h}{1-\cos h} $$
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