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Chapter 4: Problem 10

The function defined by \(f(x)=x^{2}-9\) (is/is not) a one-to-one function, whereas \(g(x)=x^{2}-9 ; x \geq 0\) (is/is not) a one-to-one function.

Short Answer

Expert verified
The function f(x) is not a one-to-one function, whereas the function g(x) is a one-to-one function.

Step by step solution

01

- Define a One-to-One Function

A function is one-to-one if every element of the range is mapped from a distinct element of the domain. In other words, if f(a) = f(b) implies that a = b, then the function is one-to-one.
02

- Analyze the Function f(x)

Consider the function f(x) = x^2 - 9. To determine if it is one-to-one, let's see if there are distinct x-values (a and b) such that f(a) = f(b).
03

- Test for One-to-One with Function f(x)

Assume f(a) = f(b). Therefore, a^2 - 9 = b^2 - 9.
04

- Simplify the Equation for f(x)

Simplifying the equation a^2 - 9 = b^2 - 9, we get a^2 = b^2. This further simplifies to a = b or a = -b. Since a is not necessarily equal to b, f(x) is not one-to-one.
05

- Analyze the Function g(x)

Consider the function g(x) = x^2 - 9 with the domain restriction x 鈮� 0. Again, let's see if distinct x-values (a and b) exist such that g(a) = g(b).
06

- Test for One-to-One with Function g(x)

Assume g(a) = g(b). Therefore, a^2 - 9 = b^2 - 9. This simplifies to a^2 = b^2. Given the domain restriction x 鈮� 0, the solutions are a = b.
07

- Conclusion for g(x)

Since the domain restriction leads to a = b only, g(x) is a one-to-one function.

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Key Concepts

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

Function Analysis
A function analysis consists of examining properties like one-to-one correspondence, injectiveness, range, and domain. For a function to be one-to-one, each output must map to exactly one unique input.

Simply put, if you have two different inputs 'a' and 'b', then the function values at 'a' and 'b' should not be equal unless 'a' equals 'b'.

In mathematical terms, a function is one-to-one (or injective) if and only if:
  • if \(f(a)=f(b)\), then \(a=b\)

This is very important in various branches of mathematics because one-to-one functions have unique inverses, making solver problems easier.
Domain Restriction
A domain restriction changes the set of inputs for which a function is defined. This can often make a function meet specific criteria, such as becoming one-to-one.

For instance, consider the function \(f(x) = x^2 - 9\). This function is not one-to-one over all real numbers because both positive and negative values of 'x' can give the same 'f(x)' value.

But if we restrict the domain to \( x \geq 0 \, \), the function's behavior changes. Now there's no negative value to map to 'f(x)', simplifying our steps:
  • \(y = x^2 - 9 \)
  • \(x = \sqrt{y + 9} \)

Now each 'y' has a unique 'x', making 'f(x)' one-to-one when the domain is restricted.
Algebraic Functions
Algebraic functions are mathematical expressions that use algebraic operations like addition, subtraction, multiplication, division, and exponentiation with variables.

For example, the function \(f(x) = x^2 - 9 \) is an algebraic function involving squaring and subtraction.
Analyzing such functions for properties like one-to-one correspondence often involves:
  • Setting \(f(a) = f(b)\)
  • Simplifying the resultant equation
  • Checking if 'a' must equal 'b'

For \(f(x) = x^2 - 9 \), we solved \(a^2 - 9 = b^2 - 9 \ \rightarrow a^2 = b^2 \rightarrow a = b \) or \(a = -b\). Since \'a\' is not necessarily equal to 'b', \(f(x) \) is not one-to-one.
The power of algebraic manipulation lies in simplifying complex expressions, hence helping to understand functions deeply.

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Most popular questions from this chapter

a. The populations of two countries are given for January 1,2000 , and for January 1,2010 . Write a function of the form \(P(t)=P_{0} e^{k t}\) to model each population \(P(t)\) (in millions) \(t\) years after January 1, 2000.$$ \begin{array}{|l|c|c|c|} \hline & \begin{array}{c} \text { Population } \\ \text { in 2000 } \\ \text { (millions) } \end{array} & \begin{array}{c} \text { Population } \\ \text { in 2010 } \\ \text { (millions) } \end{array} & \boldsymbol{P}(t)=\boldsymbol{P}_{0} e^{k t} \\ \hline \text { Switzerland } & 7.3 & 7.8 & \\ \hline \text { Israel } & 6.7 & 7.7 & \\ \hline \end{array}$$ b. Use the models from part (a) to predict the population on January \(1,2020,\) for each country. Round to the nearest hundred thousand. c. Israel had fewer people than Switzerland in the year 2000 , yet from the result of part (b), Israel will have more people in the year \(2020 ?\) Why? d. Use the models from part (a) to predict the year during which each population will reach 10 million if this trend continues.

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(6 \log _{5}(4 p-3)=18\)

Write the domain in interval notation. $$ f(x)=\log _{5}(5-3 x) $$

Determine if the statement is true or false. For each false statement, provide a counterexample. For example, \(\log (x+y) \neq \log x+\log y\) because \(\log (2+8) \neq \log 2+\log 8\) (the left side is 1 and the right side is approximately 1.204 ). $$ \log (x y)=(\log x)(\log y) $$

A table of data is given. a. Graph the points and from visual inspection, select the model that would best fit the data. Choose from $$\begin{array}{ll} y=m x+b \text { (linear) } & y=a b^{x} \text { (exponential) } \\ y=a+b \ln x \text { (logarithmic) } & y=\frac{c}{1+a e^{-b x}} \text { (logistic) } \end{array}$$ b. Use a graphing utility to find a function that fits the data. $$ \begin{array}{|c|c|} \hline x & y \\ \hline 3 & 2.7 \\ \hline 7 & 12.2 \\ \hline 13 & 25.7 \\ \hline 15 & 30 \\ \hline 17 & 34 \\ \hline 21 & 44.4 \\ \hline \end{array} $$
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