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Chapter 8: Problem 61

Explain how to identify the domain and range of a function from its graph.

Short Answer

Expert verified
The domain of a function is the set of all input values (x-values on the graph), and the range is the set of all output values (y-values on the graph). The range and domain can be identified by looking at the vertical and horizontal extents of the graph respectively.

Step by step solution

01

Understand the terminologies

The 'domain' of a function refers to all possible input values, which are usually represented by 'x' on the graph. The 'range' represents all possible output values that are determined by the function, usually represented by 'y' on the graph.
02

Identify the Domain

Looking at the horizontal extent of the graph, determine the smallest and largest x-values. The domain will be all x-values between these two extremes, inclusive. If the graph extends infinitely to the left or right, then the domain may be all real numbers.
03

Identify the Range

Now, observe the vertical extent of the graph to identify the range. Check for the smallest and largest y-values. The range will be all y-values between these two points, inclusive. If the graph continues infinitely upwards or downwards, then the range could be all real numbers.

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Most popular questions from this chapter

Simplify: \(3\left(\frac{x-2}{3}\right)+2\)

Explain how to determine if two functions are inverses of each other.

The functions are all one-to-one. For each function, a. Find an equation for \(f^{-1}(x),\) the inverse function. b. Verify that your equation is correct by showing that \(f\left(f^{-1}(x)\right)=x\) and \(f^{-1}(f(x))=x\). $$f(x)=\frac{2 x-3}{x+1}$$

Use a graphing utility to graph \(f\) and \(g\) in the same viewing rectangle. In addition, graph the line \(y=x\) and visually determine if \(f\) and g are inverses. $$f(x)=\sqrt[3]{x}-2, \quad g(x)=(x+2)^{3}$$

Let $$\begin{array}{l}f(x)=2 x-5 \\\g(x)=4 x-1 \\\h(x)=x^{2}+x+2\end{array}$$. Evaluate the indicated function without finding an equation for the function. $$g(f[h(1)])$$
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