/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 116 Determine whether each statement... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine whether each statement is true or false. For \(y=x^{2}, y\) is a function of \(x\)

Short Answer

Expert verified
True, \(y = x^2\) is a function of \(x\) as each \(x\) has a unique \(y\).

Step by step solution

01

Understand the Definition of a Function

A function is a relation where each input has exactly one output. In mathematical terms, for every value of the independent variable (often called "input"), there is one and only one value of the dependent variable (often called "output"). This is usually written as \(y = f(x)\).
02

Analyze the Given Equation

We have the equation \(y = x^2\). This expresses \(y\) as a formula in terms of \(x\). Here, \(x\) is the independent variable and \(y\) is the dependent variable.
03

Check the Uniqueness of Outputs

Substitute different values of \(x\) into \(y = x^2\). For example, if \(x = 2\), then \(y = 2^2 = 4\). If \(x = -2\), then \(y = (-2)^2 = 4\) as well. Each specific value of \(x\) results in just one value of \(y\).
04

Conclude if Relation is a Function

Since for every value of \(x\) there is exactly one value of \(y\), \(y = x^2\) satisfies the condition of a function. Thus, \(y\) is a function of \(x\).

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Key Concepts

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

Dependent Variable
In mathematics, when discussing functions, the concept of a dependent variable is incredibly important. The dependent variable is the output or result of a function, which depends on the input value or values provided to the function. Imagine you have a recipe; the final dish you make is the dependent variable that changes based on the ingredients you use.
  • In the equation \(y = x^2\), \(y\) is the dependent variable.
  • It represents the output of the function, which changes according to the value squared of the independent variable, \(x\).
  • Every value of \(x\) leads to a single output or value of \(y\), making \(y\) truly dependent on the specific value of \(x\).
In the example of the function \(y = x^2\), when \(x\) is changed, like from \(2\) to \(-2\), \(y\) results in the same output \(4\) because squaring both values results in \(4\). This demonstrates how the dependent variable relies on the independent variable's specific value.
Independent Variable
The independent variable is the core component in functions that determines the value of the dependent variable. It is often likened to an ingredient you choose freely, which then influences the outcome, or the dependent variable, in your mathematical 'recipe'.
  • In the function \(y = x^2\), \(x\) acts as the independent variable.
  • It is the value you choose, and then substitute into the function, affecting the result or output.
  • The flexibility of choosing \(x\)'s value differentiates it from the dependent variable, which adjusts accordingly.
A valuable way to think about the independent variable is imagining it as the initial condition or setup. By altering \(x\), you decide what the function will calculate, and you directly observe how changes in \(x\) affect \(y\). For example, when choosing \(x = 3\) or \(x = -3\), you calculate \(y = 9\) in both cases, showcasing how your choice of \(x\) independently influences the outcome.
Function Definition
Understanding the precise definition of a function is essential in mathematics. A function is essentially a special type of relation. It uniquely matches each input to exactly one output, ensuring consistency and predictability.
  • This means that for every independent variable value, there should be one and only one dependent variable outcome.
  • The specific arrangement is often expressed in notation like \(y = f(x)\).
  • In \(y = x^2\), for any chosen \(x\), there's a corresponding \(y\) value derived using the function's rule.
A crucial characteristic of functions is this one-to-one relationship between inputs and outputs. If you analyze \(y = x^2\), you'll see that each value plugged in for \(x\) produces a distinct and singular \(y\). This adheres to the function's definition strictly, proving \(y\) is indeed a function of \(x\). It is these predictable outcomes that allow functions to be so useful in various fields, representing real-world processes where inputs and outputs are consistently linked.

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Most popular questions from this chapter

In Exercises \(51-60\), show that \(f(g(x))=x\) and \(g(f(x))=x\). $$f(x)=\frac{x-2}{3}, \quad g(x)=3 x+2$$

Typical supply and demand relationships state that as the number of units for sale increases, the market price decreases. Assume that the market price \(p\) and the number of units for sale \(x\) are related by the demand equation: $$p=10,000-\frac{1}{4} x$$ Assume that the cost \(C(x)\) of producing \(x\) items is governed by the equation $$C(x)=30,000+5 x$$ and the revenue \(R(x)\) generated by selling \(x\) units is governed by $$R(x)=1000 x$$ a. Write the cost as a function of price \(p\) b. Write the revenue as a function of price \(p\) c. Write the profit as a function of price \(p\)

Let \(f(x)=x^{2}+1 .\) Graph \(y_{1}=f(x)\) and \(y_{2}=f(x-2)\) in the same viewing window. Describe how the graph of \(y_{2}\) can be obtained from the graph of \(y_{1}\)

In Exercises \(77-81,\) for the functions \(f(x)=x+2\) and \(g(x)=x^{2}-4,\) find the indicated function and state its domain. Explain the mistake that is made in each problem. $$f \circ g$$ Solution: $$\begin{aligned}f \circ g &=f(x) g(x) \\\&=(x+2)\left(x^{2}-4\right) \\\&=x^{3}+2 x^{2}-4 x-8\end{aligned}$$ Domain: \((-\infty, \infty)\) This is incorrect. What mistake was made?

Refer to the following: In calculus, the difference quotient \(\frac{f(x+h)-f(x)}{h}\) of a function fis used to find the derivative \(f^{\prime}\) of \(f\), by allowing \(h\) to approach zero, \(h \rightarrow 0 .\) The derivative of the inverse function \(\left(f^{-1}\right)^{\prime}\) can be found using the formula $$\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}$$ provided that the denominator is not 0 and both \(f\) and \(f^{-1}\) are differentiable. For the following one-to-one function, find (a) \(f^{-1},\) (b) \(f^{\prime},\) (c) \(\left(f^{-1}\right)^{\prime},\) and (d) verify the formula above. For (b) and (c), use the difference quotient. $$f(x)=2 x+1$$

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