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91Ó°ÊÓ

Determine the intervals on which \(f(x)\) is continuous. $$f(x)=\sqrt{x+3}$$

Short Answer

Expert verified
The function \(f(x) = \sqrt{x + 3}\) is continuous on the interval \([-3, \infty)\).

Step by step solution

01

Identify the Argument of the Square Root

In the function \(f(x) = \sqrt{x + 3}\), the argument of the square root is \(x + 3\).
02

Determine Values where the Argument is Non-Negative

To be defined, the argument of the square root must be non-negative. Hence, we set \(x + 3 \geq 0\). Solving this inequality for \(x\) gives \(x \geq -3\). That is, the function \(f(x) = \sqrt{x + 3}\) is defined for all \(x \geq -3\).
03

Conclude the Intervals of Continuity

Since \(f(x) = \sqrt{x + 3}\) is a single-valued function that is defined for all \(x \geq -3\), it is continuous for \(x \geq -3\). So, the interval of continuity is \([-3, \infty)\).

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Key Concepts

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

Interval of Continuity
Continuity of a function is a fundamental concept in calculus. It refers to a function that has no "gaps," "breaks," or "jumps" along its graph. When it comes to determining the interval of continuity, we need to find where these gaps do not occur.
For the function given in the problem, \(f(x) = \sqrt{x+3}\), we need to find where this function is defined and continuous. A function is continuous at an interval if it is defined on that interval and there are no sudden jumps or breaks.
Since the square root function is only defined for non-negative numbers, we must ensure the argument inside the square root (\(x + 3\)) does not become negative. This leads us to solve the inequality \(x + 3 \geq 0\). Solving it gives \(x \geq -3\), thus defining the function for every \(x\) greater than or equal to \(-3\).
  • The interval for which \(f(x)\) is continuous: \([-3, \infty)\)
This means our function is without interruption, just like a smooth path, for all \(x\) values from \(-3\) to infinity.
Functions
In mathematics, a function is like a machine that takes an input and gives back exactly one output. It relates each element of a set with exactly one element of another set, often seen through simple equations. Functions are key in solving real-world problems by using mathematical relationships.
Our particular function in the exercise is \(f(x) = \sqrt{x+3}\). This is a type of function known as a square root function. It takes the form \(\sqrt{u}\), where \(u = x + 3\). Here, each input \(x\) produces exactly one output, ensuring it adheres to the rule of a function.
  • A function assigns one output for each input
  • \(f(x) = \sqrt{x+3}\) is defined for \(x \geq -3\)
This function maps every possible input \(x\) from \(-3\) onwards to a non-negative number, reflecting the essence of continuity in functions.
Inequalities
Inequalities are expressions involving the symbols \(>\), \(<\), \(\geq\), or \(\leq\). They provide us with a range of values rather than a single value solution. Solving inequalities is critical when it comes to defining functions where arguments must satisfy certain conditions.
For the function \(f(x) = \sqrt{x+3}\), the argument inside the square root must be non-negative for the function to exist. Therefore, we examine the inequality \(x + 3 \geq 0\). It is solved by simplifying or accepting all values that make the inequality true.
  • Subtract 3 from both sides, yielding \(x \geq -3\)
This inequality tells us the condition required for our function to remain valid and continuous. The solution, \(x \geq -3\), provides the range of values for which the function can be evaluated without producing undefined results. Understanding inequalities helps in acknowledging these conditions and determining where a function can smoothly transition through its domain.

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Most popular questions from this chapter

A function is continuous from the right at \(x=a\) if \(\lim _{x \rightarrow a^{+}} f(x)=f(a) .\)Determine whether \(f(x)\) is continuous from the right at \(x=2.\) $$f(x)=\left\\{\begin{array}{ll} x^{2} & \text { if } x<2 \\ 3 x-2 & \text { if } x>2 \end{array}\right.$$

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