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At what points are the functions in Exercises \(13-30\) continuous? $$y=\sqrt{2 x+3}$$

Short Answer

Expert verified
The function is continuous for \( x \geq -\frac{3}{2} \).

Step by step solution

01

Identify the Function's Continuity Requirements

The function is given by \( y = \sqrt{2x + 3} \). A square root function is continuous for all values where the expression inside the square root is non-negative. Therefore, the expression \( 2x + 3 \) must be greater than or equal to 0.
02

Solve the Inequality for Continuity

Set up the inequality from the condition identified:\[ 2x + 3 \geq 0 \]Subtract 3 from both sides:\[ 2x \geq -3 \]Divide by 2:\[ x \geq -\frac{3}{2} \]
03

Conclusion on Continuity Points

The solution \( x \geq -\frac{3}{2} \) indicates that the function is continuous for all \( x \) values greater than or equal to \(-\frac{3}{2}\). Therefore, the function \( y = \sqrt{2x + 3} \) is continuous on the interval \([-\frac{3}{2}, \infty)\).

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Key Concepts

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

Square Root Functions
Square root functions are a type of function involving a radical expression. Here, our function is given as \( y = \sqrt{2x + 3} \). The square root function is defined only for values where the radicand, which is the expression inside the square root, is non-negative. This is because in real numbers, you cannot take the square root of a negative number without venturing into complex numbers.
For our function, the radicand is \( 2x + 3 \). Hence, to find where the function is continuous, we need it to be non-negative.
This means the expression \( 2x + 3 \) must be greater than or equal to zero. Solving this will determine the domain of the function, which also points to its continuity.
Inequalities
Inequalities are mathematical expressions involving the symbols \(<, >, \leq, \geq\). They are essential in determining the range of values for which the expression holds true. In our problem, the inequality \( 2x + 3 \geq 0 \) defines where the function \( y = \sqrt{2x + 3} \) is valid.
By solving this inequality -
  • First, subtract 3 from both sides to get \( 2x \geq -3 \).
  • Then, divide by 2 to isolate \( x \), resulting in \( x \geq -\frac{3}{2} \).

This solution tells us that the function is continuous for all \( x \) values greater than or equal to \(-\frac{3}{2}\). This approach ensures that the function remains within the set of real numbers.
Calculus Continuity Concepts
In calculus, continuity of a function at a certain point means that there are no breaks, jumps, or holes at that point for the function. For a function \( f(x) \) to be continuous at a point \( c \), it must satisfy three main conditions:
  • \( f(c) \) must be defined.
  • The limit of \( f(x) \) as \( x \) approaches \( c \) must exist.
  • The limit of \( f(x) \) as \( x \) approaches \( c \) must be equal to \( f(c) \).
For the square root function \( y = \sqrt{2x + 3} \), we test continuity over the interval \([-\frac{3}{2}, \infty)\). At every point in this interval, the function satisfies all the above criteria. As such, the function has no interruptions or discontinuities within this range. Understanding these concepts helps students determine not just where functions are defined, but also where they are smooth and unbroken.

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Most popular questions from this chapter

You will find a graphing calculator useful for Exercise. Let \(g(\theta)=(\sin \theta) / \theta.\) a. Make a table of the values of \(g\) at values of \(\theta\) that approach \(\theta_{0}=0\) from above and below. Then estimate \(\lim _{\theta \rightarrow 0} g(\theta).\) b. Support your conclusion in part (a) by graphing \(g\) near \(\theta_{0}=0.\)

Graph the functions.Then answer the following questions. a. How does the graph behave as \(x \rightarrow 0^{+} ?\) b. How does the graph behave as \(x \rightarrow \pm \infty ?\) c. How does the graph behave near \(x=1\) and \(x=-1 ?\) Give reasons for your answers. $$y=\frac{3}{2}\left(x-\frac{1}{x}\right)^{2 / 3}$$

Gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon \geq 0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$f(x)=m x, \quad m>0, \quad L=2 m, \quad c=2, \quad \epsilon=0.03$$

Use a CAS to perform the following steps: a. Plot the function \(y=f(x)\) near the point \(c\) being approached. b. Guess the value of the limit \(L\) and then evaluate the limit symbolically to see if you guessed correctly. c. Using the value \(\epsilon=0.2,\) graph the banding lines \(y_{1}=L-\epsilon\) and \(y_{2}=L+\epsilon\) together with the function \(f\) near \(c\) d. From your graph in part (c), estimate a \(\delta>0\) such that for all \(x\) \(0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon\) Test your estimate by plotting \(f, y_{1},\) and \(y_{2}\) over the interval \(0<|x-c|<\delta .\) For your viewing window use \(c-2 \delta \leq\) \(x \leq c+2 \delta\) and \(L-2 \epsilon \leq y \leq L+2 \epsilon .\) If any function values lie outside the interval \([L-\epsilon, L+\epsilon],\) your choice of \(\delta\) was too large. Try again with a smaller estimate. e. Repeat parts (c) and (d) successively for \(\epsilon=0.1,0.05,\) and 0.001 $$f(x)=\frac{\sin 2 x}{3 x}, \quad c=0$$

Give an example of a function \(g(x)\) that is continuous for all values of \(x\) except \(x=-1,\) where it has a nonremovable discontinuity. Explain how you know that \(g\) is discontinuous there and why the discontinuity is not removable.

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