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Chapter 1: Problem 2

For the function $$f(x)=\left\\{\begin{array}{ll} 2 x-5, & 0 \leq x<3 \\ 4, & x=3 \\ x^{2}-6 x+10, & 3

Short Answer

Expert verified
The limits are: \(\lim_{x \rightarrow 3^{+}} f(x) = 1\), \(\lim_{x \rightarrow 3^{-}} f(x) = 1\), \(\lim_{x \rightarrow 3} f(x) = 1\).

Step by step solution

01

Understand the Function

The function given is piecewise defined with three different expressions depending on the value of x. Recognize the different parts of the function and the intervals they apply to: 1. For \(0 \leq x < 3\), the function is \(f(x) = 2x - 5\). 2. For \(x = 3\), the function is \(f(x) = 4\). 3. For \(x > 3\), the function is \(f(x) = x^{2} - 6x + 10\).
02

Find the Right-Sided Limit

To find \(\lim_{x \rightarrow 3^{+}} f(x)\), use the expression for \(f(x)\) when \(x > 3\) since approaching 3 from the right falls into this interval. Calculate the limit: \(f(x) = x^{2} - 6x + 10\) When \(x \rightarrow 3^{+}\), substitute \(3\) into the expression: \[ f(3) = 3^{2} - 6(3) + 10 = 9 - 18 + 10 = 1 \Rightarrow \lim_{x \rightarrow 3^{+}} f(x) = 1 \]
03

Find the Left-Sided Limit

To find \(\lim_{x \rightarrow 3^{-}} f(x)\), use the expression for \(f(x)\) when \(0 \leq x < 3\) since approaching 3 from the left falls into this interval. Calculate the limit: \(f(x) = 2x - 5\) When \(x \rightarrow 3^{-}\), substitute \(3\) into the expression: \[ f(3) = 2(3) - 5 = 6 - 5 = 1 \Rightarrow \lim_{x \rightarrow 3^{-}} f(x) = 1 \]
04

Find the Two-Sided Limit

To find \(\lim_{x \rightarrow 3} f(x)\), compare the right-sided and left-sided limits. Since \(\lim_{x \rightarrow 3^{+}} f(x) = 1\) and \(\lim_{x \rightarrow 3^{-}} f(x) = 1\), the two-sided limit exists and is the same at 1: \[ \lim_{x \rightarrow 3} f(x) = 1 \]

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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 defined by different expressions depending on the value of the variable in various intervals. It allows a function to behave differently in different sections of its domain.

In the given exercise, the function is defined in three pieces:
  • For \(0 \leq x < 3\), the function is \(f(x) = 2x - 5\).
  • For \(x = 3\), the function is \(f(x) = 4\).
  • For \(x > 3\), the function is \(f(x) = x^{2} - 6x + 10\).

Each piece applies to a different interval of the variable \(x\). Understanding the breakdown of the intervals is crucial for solving limits and analyzing continuity.
One-Sided Limits
One-sided limits consider the value of a function as the variable approaches from either the left or right side of a specified point.

For example:
  • The right-sided limit, denoted as \(\lim_{{x \to c^+}} f(x)\), examines \(x\) as it approaches \(c\) from the right (values greater than \(c\)).
  • The left-sided limit, denoted as \(\lim_{{x \to c^-}} f(x)\), examines \(x\) as it approaches \(c\) from the left (values less than \(c\)).

In our exercise:

To find \(\lim_{{x \to 3^+}} f(x)\), use the expression for \(x > 3\):\(f(x) = x^{2} - 6x + 10\). When \(x \to 3^+\), we have:
\[f(3) = 3^{2} - 6(3) + 10 = 1 \Rightarrow \lim_{{x \to 3^+}} f(x) = 1\]
For \(\lim_{{x \to 3^-}} f(x)\), use the expression for \(0 \leq x < 3\):\(f(x) = 2x - 5\). When \(x \to 3^-\), we have:
\[f(3) = 2(3) - 5 = 1 \Rightarrow \lim_{{x \to 3^-}} f(x) = 1\]
Two-Sided Limits
A two-sided limit is analyzed by evaluating both the left-sided and right-sided limits at a particular point.

If both one-sided limits exist and are equal, then the two-sided limit at that point also exists and is equal to that common value.

In our exercise, since:
  • \(\lim_{{x \to 3^+}} f(x) = 1\)
  • \(\lim_{{x \to 3^-}} f(x) = 1\)

The two-sided limit exists at \(x = 3\) and \

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Most popular questions from this chapter

For a certain function \(f\), its derivative is known to be \(f^{\prime}(x)=(x-1) e^{-x^{2}}\). Note that you do not know a formula for \(y=f(x)\). a. At what \(x\) -value(s) is \(f^{\prime}(x)=0\) ? Justify your answer algebraically, but include a graph of \(f^{\prime}\) to support your conclusion. b. Reasoning graphically, for what intervals of \(x\) -values is \(f^{\prime \prime}(x)>0\) ? What does this tell you about the behavior of the original function \(f ?\) Explain. c. Assuming that \(f(2)=-3\), estimate the value of \(f(1.88)\) by finding and using the tangent line approximation to \(f\) at \(x=2\). Is your estimate larger or smaller than the true value of \(f(1.88) ?\) Justify your answer.

Suppose that \(f(x)\) is a function with \(f(125)=76\) and \(f^{\prime}(125)=8\). Estimate \(f(128.5)\). \(f(128.5)=\) \(\square\)

The cost, \(C\) (in dollars) to produce \(g\) gallons of ice cream can be expressed as \(C=f(g)\). (a) In the expression \(f(300)=350\), what are the units of \(300 ?\) [Choose: ? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] what are the units of \(350 ?\) [Choose: ? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] (b) In the expression \(f^{\prime}(300)=1.2,\) what are the units of 300? [Choose:? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] what are the units of 1.2? [Choose: ? | dollars | gallons | dollars*gallons | dollars/gallon | gallons/dollar] (Be sure that you can carefully put into words the meanings of each of these statement in terms of ice cream and money.)

A bungee jumper dives from a tower at time \(t=0\). Her height \(s\) in feet at time \(t\) in seconds is given by \(s(t)=100 \cos (0.75 t) \cdot e^{-0.2 t}+100\). a. Write an expression for the average velocity of the bungee jumper on the interval \([1,1+\) \(h]\) b. Use computing technology to estimate the value of the limit as \(h \rightarrow 0\) of the quantity you found in (a). c. What is the meaning of the value of the limit in (b)? What are its units?

The displacement (in meters) of a particle moving in a straight line is given by $$s=t^{2}-5 t+12$$ where \(t\) is measured in seconds. (A) (i) Find the average velocity over the time interval [3,4] . Average Velocity \(=\square\) (ii) Find the average velocity over the time interval [3.5,4]. Average Velocity \(=\square\) (iii) Find the average velocity over the time interval [4,5] . Average Velocity \(=\square\) (iv) Find the average velocity over the time interval [4,4.5] . Average Velocity \(=\square\) (B) Find the instantaneous velocity when \(t=4\) \(=\square\) Instantaneous velocity \(=\square\)
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