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Chapter 1: Problem 57

Find the domain of the function. $$f(x)=5 x^{2}+2 x-1$$

Short Answer

Expert verified
The domain of the function \(f(x) = 5x^{2} + 2x - 1\) is all real numbers.

Step by step solution

01

Recognize Type of Function

Identify the given function as a quadratic function. A quadratic function is a function that can be described by an equation of the form \(f(x) = ax^{2} + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a 鈮� 0\).
02

Domain of a Quadratic Function

Understand that the domain of a quadratic function is all real numbers. This is because there is no real number that we can square (for the \(x^{2}\) part) that will result in a number that is not real or that is undefined.
03

Apply to Given Function

Therefore, the domain of the function \(f(x) = 5x^{2} + 2x - 1\) is all real numbers, as there are no restrictions on the values for x in the given function.

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Most popular questions from this chapter

You can encode and decode messages using functions and their inverses. To code a message, first translate the letters to numbers using 1 for "A," 2 for "B," and so on. Use 0 for a space. So, "A ball" becomes 1 0 2 1 12 12. Then, use a one-to-one function to convert to coded numbers. Using \(f(x)=2 x-1,\) "A ball" becomes 1 ?1 3 1 23 23. (a) Encode "Call me later" using the function \(f(x)=5 x+4.\) (b) Find the inverse function of \(f(x)=5 x+4\) and use it to decode 119 44 9 104 4 104 49 69 29.

Perform the operation and simplify. $$\frac{x+7}{2(x-9)} \div \frac{x-7}{2(x-9)}$$

Determine whether the statement is true or false. Justify your answer.The graphs of \(f(x)=|x|+6\) and \(f(x)=|-x|+6\) are identical.

Determine whether the function is even, odd, or neither (a) algebraically, (b) graphically by using a graphing utility to graph the function, and (c) numerically by using the table feature of the graphing utility to compare \(f(x)\) and \(f(-x)\) for several values of \(x\). $$g(x)=x^{3}-5 x$$

Use the functions \(f(x)=x+4\) and \(g(x)=2 x-5\) to find the specified function. $$g^{-1} \circ f^{-1}$$
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