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Chapter 2: Problem 103

If a function is defined by an equation, explain how to find its domain.

Short Answer

Expert verified
To find the domain of a function defined by an equation, one would identify potential restrictions on the domain due to operations like division (averting division by zero), square roots (the radicand must be equal to or greater than zero), and logarithms (the argument must be more than zero). By resolving these restrictions, the non-allowable values are found; all other values form the domain of the function.

Step by step solution

01

Understand the Concept of a Function's Domain

First, it should be noted that the domain of a function is the set of all 'allowable' input values or arguments for which the function is defined. In other words, these are the values that can be substituted into the function to obtain a real, valid output. In the case of functions defined by equations, these input values are usually represented by the variable \(x\). So, the first step is being clear on what a function's domain entails.
02

Identify Possible Restrictions on the Domain

The next step is identifying possible factors or operations in the equation that could restrict the set of allowable input values. Three common operations that can limit a domain include: 1) Division: The domain will exclude any \(x\) value that makes the denominator zero. 2) Square roots (or any even-root operation) : The expression under the square root must be equal to or greater than zero. 3) Logarithms: The log's argument must be more than zero. Each operation's restriction should be identified.
03

Resolve the Restrictions

After identifying the restrictions on the domain from the second step, the next step is resolving these restrictions to find the values that are not included in the domain. For division, this means solving the denominator equation for \(x\) when it equals zero. For square roots, it involves solving the radicand (the expression under the root) equation for when it's less than zero. And logarithmic restrictions require solving the logarithm's argument for when it's equal to or less than zero. With these solutions for \(x\), the restrictions on the domain are resolved.
04

Combine All Allowable Values

Finally, after finding the restrictions on the domain and the corresponding non-allowable \(x\) values, the rest of the \(x\) values are part of the domain. This may involve forming intervals for the allowable values, especially when the function is defined over the set of real numbers. Consequently, the domain of the function is established.

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Most popular questions from this chapter

give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation's domain and range. $$ x^{2}+y^{2}=16 $$

complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}+12 x-6 y-4=0 $$

give the center and radius of the circle described by the equation and graph each equation. Use the graph to identify the relation's domain and range. $$ (x+2)^{2}+y^{2}=16 $$

Define a piecewise function on the intervals \((-\infty, 2],(2,5)\) and \([5, \infty)\) that does not "jump" at 2 or 5 such that one piece is a constant function, another piece is an increasing function, and the third piece is a decreasing function.

determine whether each statement makes sense or does not make sense, and explain your reasoning. The graph of \((x-3)^{2}+(y+5)^{2}=-36\) is a circle with radius 6 centered at \((3,-5)\)
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