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Chapter 3: Problem 1

Let \(f(x)=x^{2}\) and find \(f(1), f(2), f(3)\) \(f(-1), f(-2)\) and \(f(-3)\). What type, one \(\rightarrow\) one or many \(\rightarrow\) one, of function is \(f(x) ?\)

Short Answer

Expert verified
\(f(1)=1, f(2)=4, f(3)=9, f(-1)=1, f(-2)=4, f(-3)=9\). The function is many-to-one.

Step by step solution

01

- Understand the function

The function given is \(f(x) = x^2\). This is a simple quadratic function where you square the value of x.
02

- Calculate \(f(1)\)

Plug in 1 for x in the function: \(f(1) = 1^2 = 1\).
03

- Calculate \(f(2)\)

Plug in 2 for x in the function: \(f(2) = 2^2 = 4\).
04

- Calculate \(f(3)\)

Plug in 3 for x in the function: \(f(3) = 3^2 = 9\).
05

- Calculate \(f(-1)\)

Plug in -1 for x in the function: \(f(-1) = (-1)^2 = 1\).
06

- Calculate \(f(-2)\)

Plug in -2 for x in the function: \(f(-2) = (-2)^2 = 4\).
07

- Calculate \(f(-3)\)

Plug in -3 for x in the function: \(f(-3) = (-3)^2 = 9\).
08

- Determine the type of function

Given the results: \(f(1) = 1\), \(f(2) = 4\), \(f(3) = 9\), \(f(-1) = 1\), \(f(-2) = 4\), and \(f(-3) = 9\), observe that different values of x can produce the same value for \(f(x)\). Therefore, \(f(x)\) is a many-to-one function.

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Key Concepts

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

Function Evaluation
Function evaluation is the process of finding the output of a function based on an input. In our example, we have a quadratic function given by \(f(x) = x^2\). To evaluate this function for a given value, we simply substitute the input value (x) into the function and perform the necessary calculations.
For instance:
  • For \(x = 1\), \(f(1) = 1^2 = 1\).
  • For \(x = 2\), \(f(2) = 2^2 = 4\).
  • For \(x =3\), \(f(3) = 3^2 = 9\).
The output of the function changes based on different inputs. Similarly, for negative values:
  • For \(x = -1\), \(f(-1) = (-1)^2 = 1\).
  • For \(x = -2\), \(f(-2) = (-2)^2 = 4\).
  • For \(x = -3\), \(f(-3) = (-3)^2 = 9\).
Understanding this basic principle of function evaluation is fundamental for solving more complex mathematical problems.
Many-to-One Function
A many-to-one function is a type of function where multiple input values can produce the same output value. In our example, \(f(x) = x^2\) is a many-to-one function. This is because:
  • Both \(x = 1\) and \(x = -1\) produce the same output, which is \(f(1) = 1\) and \(f(-1) = 1\).
  • Similarly, \(x = 2\) and \(x = -2\) both yield \(f(2) = 4\) and \(f(-2) = 4\).
  • Lastly, \(x = 3\) and \(x = -3\) give the same result: \(f(3) = 9\) and \(f(-3) = 9\).
In contrast to one-to-one functions, where each input has a unique output, many-to-one functions allow for different inputs to be mapped to the same output. Understanding this concept helps in distinguishing between various types of functions in mathematics.
Mathematical Problem Solving
Problem-solving in mathematics involves a series of logical steps and the application of mathematical methods to find solutions. In our example, solving the problem of evaluating \(f(x) = x^2\) for various input values involves straightforward substitution and calculation.
Here鈥檚 how to approach such problems:
  • First, understand the given function. For \(f(x) = x^2\), recognize that the output is simply the square of the input.
  • Next, substitute the given values of \(x\) into the function. For instance, to find \(f(1)\), substitute 1 for \(x\) resulting in \(1^2 = 1\).
  • Repeat this for all provided input values. Always keep track of positive and negative values separately as they can yield the same result in many-to-one functions.
  • Finally, analyze the results to determine the nature of the function, whether one-to-one or many-to-one, as shown in our example.
Breaking down complex problems into simple, manageable steps like this enhances understanding and makes the process of mathematical problem-solving easier and more efficient.

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Most popular questions from this chapter

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if [controlengineering] The steady-state error, \(e_{\mathrm{ss}}\), of a system is given by $$ e_{5 \mathrm{~s}}=\lim _{s \rightarrow 0} \frac{M}{1+G(s)} $$ where \(M\) is a constant. Given that $$ G(s)=\left(\frac{k_{1}+k_{2}}{s}\right) \times\left(\frac{1}{1+s \tau}\right) $$ where \(k_{1}, k_{2}\) and \(\tau\) are non-zero constants, evaluate \(e_{\mathrm{ss}}\). Employ algebra to find the limits of questions \(\mathbf{8}\) and \(\mathbf{9}\).
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