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Chapter 3: Problem 79

Let \(f(x)=\left\\{\begin{array}{ll}2 x+1 & \text { if } x \leq 0 \\ x-k & \text { if } x>0\end{array}\right.\) Find \(k\) so that \(f(x)\) is continuous everywhere

Short Answer

Expert verified
The value of \(k\) for continuity is \(-1\).

Step by step solution

01

Identify point of continuity

To ensure continuity of the function everywhere, we examine the point where the piecewise function transitions, which is at \(x = 0\). We need to ensure that the left-hand limit, the right-hand limit, and the value of the function at \(x = 0\) are equal.
02

Calculate the left-hand limit

The left-hand limit of \(f(x)\) as \(x\) approaches 0 from the left is determined using the expression \(2x + 1\). So, \(\lim_{{x \to 0^-}} f(x) = 2(0) + 1 = 1\).
03

Calculate the right-hand limit

The right-hand limit of \(f(x)\) as \(x\) approaches 0 from the right is determined using the expression \(x - k\). So, \(\lim_{{x \to 0^+}} f(x) = 0 - k = -k\).
04

Evaluate the function at x=0

Since \(x = 0\) falls under \(x \leq 0\), we use the expression \(2x + 1\) to find \(f(0) = 2(0) + 1 = 1\).
05

Set limits and function value equal

For continuity at \(x = 0\), set the left-hand limit, right-hand limit, and \(f(0)\) equal to each other: \[ 1 = -k = 1 \].
06

Solve equation for k

From \(1 = -k\), solve for \(k\) by multiplying both sides by -1: \(-1 = k\).

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Key Concepts

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

Piecewise Functions
A piecewise function is like a puzzle with different rules in various parts of its domain. These functions are used to represent situations where a rule or behavior changes based on the input value. The function given in our problem is a classic piecewise function:
  • It uses the formula \(2x + 1\) when \(x \leq 0\).
  • It switches to \(x - k\) when \(x > 0\).
So, at \(x = 0\), there's a transition between the two expressions. To understand piecewise functions well, it's essential to identify and manage these boundaries where the formulas change. Think of these as special spots that might need extra care to ensure smoothness or specific properties, like continuity!
Limits in Calculus
Limits are like calculus's way of peeking at what happens "just before" or "just after" a certain point.
They are vital when dealing with piecewise functions, especially at transition points. To see how a function behaves when approaching our point of interest, we consider:
  • Left-hand limit: This is when we approach from the left, noted as \( \, \lim_{{x \to 0^-}} \, \). For our function, this is determined using \(2x + 1\), which equals 1 when \(x\) gets very close to 0 from the left.
  • Right-hand limit: This observes the behavior from the right side, denoted as \( \, \lim_{{x \to 0^+}} \, \). Here, we use \(x - k\). The right-hand limit becomes \(-k\) as \(x\) approaches 0 from the right.
Understanding limits helps ensure that a function transitions smoothly from one piece to another without jumps or breaks, which is crucial for continuity.
Solving Equations
In calculus, solving equations allows us to handle values that bring different pieces of a function together. For continuity, the left-hand limit, right-hand limit, and the function's value at the transition must be the same.
Here, we need to solve the equation:
  • The left-hand limit is 1.
  • The right-hand limit is \(-k\), which should also be 1.
  • The function value at \(x=0\) is 1.
Putting these together, we set equation: \(1 = -k = 1\). This tells us all three are equal.
To solve for \(k\), we rearrange \(-k = 1\) to get \(k = -1\).
Solving these kinds of equations is a practical skill to ensure a piecewise function remains continuous, so it behaves predictably across its entire domain.

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Most popular questions from this chapter

Is it true that if \(\lim _{x \rightarrow 0} f(x)=0\) and \(\lim _{x \rightarrow 0} g(x)=0,\) then \(\lim _{x \rightarrow 0}\left(\frac{f(x)}{g(x)}\right)\) does not exist? Explain why this is true or give an example that shows it is not true.

Find, without graphing, where each of the given functions is continuous. $$ f(x)=\left\\{\begin{array}{ll} x & \text { if } x<1 \\ 2 & \text { if } x=1 \\ x-1 & \text { if } x>1 \end{array}\right. $$

Lactation Curves Cardellino and Benson \(^{34}\) studied lactation curves for ewes rearing lambs. They created the mathematical model given approximately by the equation \(P(D)=273+5.4 D-0.13 D^{2},\) where \(D\) is day of lactation and \(P\) is milk production in grams. Graph this equation on the interval \([6,60] .\) Find the instantaneous rate of change of milk production with respect to the day of lactation when (a) \(D=10,\) (b) \(D=50 .\) Give units and interpret your answer.

Cost Curve Johnston \(^{68}\) made a statistical estimation of the cost-output relationship for 40 firms. The data for one of the firms is given in the following table. $$ \begin{array}{|c|cccccc|} \hline x & 2.5 & 3 & 4 & 5 & 6 & 7 \\ \hline y & 20 & 26 & 20 & 27 & 30 & 29 \\ \hline x & 11 & 22.5 & 29 & 37 & 43 & \\ \hline y & 42 & 53 & 75 & 73 & 78 & \\ \hline \end{array} $$ Here \(x\) is the output in millions of units, and \(y\) is the total cost in thousands of pounds sterling. a. Determine both the best-fitting line using least squares and the square of the correlation coefficient. Graph. b. Determine both the best-fitting quadratic using least squares and the square of the correlation coefficient. Graph. c. Which curve is better? Why? d. Using the quadratic cost function found in part (b) and the approximate derivative found on your computer or graphing calculator, graph the marginal cost. What is happening to marginal cost as output increases? Explain what this means.

In Exercise 67 of Section 3.3 you found that \(d / d x(\sin x)=\) \(\cos x .\) Notice from your computer or calculator that \(\sin 0=\) 0 and \(\cos 0=1\). Use all of this and the tangent line approximation to approximate \(\sin 0.02\). Compare your answer to the value your computer or calculator gives.
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