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Say whether or not \(f(x)\) is defined for the given values of \(x .\) If it is defined, give its value. \(f(x)=x-\frac{1}{x^{2}}\), with domain \((0,+\infty)\) a. \(x=4\) b. \(x=0\) c. \(x=-1\)

Short Answer

Expert verified
The function \(f(x)=x-\frac{1}{x^{2}}\) is defined for \(x=4\) with a value of \(\frac{63}{16}\) but is not defined for \(x=0\) and \(x=-1\), as they are not within the domain of \((0, +\infty)\).

Step by step solution

01

Check if \(x=4\) is in the domain

The domain for the function is given as \((0, +\infty)\). Since 4 is greater than 0 and within the domain of the function, the function is defined for \(x=4\).
02

Calculate the function value for \(x=4\)

Since we found that the function is defined for \(x=4\), we can now calculate the value of the function at this point. Plug in \(x=4\) into the function, \(f(x)=x-\frac{1}{x^{2}}\): \( f(4)=4-\frac{1}{4^{2}} \) \( f(4)=4-\frac{1}{16} \) \( f(4)=\frac{63}{16} \)
03

Check if \(x=0\) is in the domain

The domain for the function is given as \((0, +\infty)\). Since 0 is not within the domain (0 is not greater than 0), the function is not defined for \(x=0\).
04

Check if \(x=-1\) is in the domain

The domain for the function is given as \((0, +\infty)\). Since -1 is smaller than 0, the function is not defined for \(x=-1\). So the function is only defined for \(x=4\) and its value at this point is \(\frac{63}{16}\).

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Key Concepts

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

Function Domain
The function domain is the set of all possible input values (often represented as \(x\)) for which the function is defined. When we talk about the function \(f(x)=x-\frac{1}{x^{2}}\), understanding where \(x\) can take values from is crucial. In this case, the domain is given as \((0, +\infty)\).
This means:
  • \(x\) can be any positive number.
  • \(x\) cannot be zero or a negative number.
This is because within the function formula, \(x^{2}\) is in the denominator, and having a value of zero would make the expression undefined due to division by zero. Hence, only values greater than zero are included, resulting in the domain \((0, +\infty)\). This simple yet fundamental step helps determine where the function can work correctly.
Function Evaluation
Function evaluation involves substituting a particular value of \(x\) into the function to calculate its output. This process tells us the value of the function at specific points. Using the given function \(f(x)=x-\frac{1}{x^{2}}\), evaluating the function for \(x = 4\), as shown, involves substituting and simplifying:
  • Substituting gives \(f(4) = 4-\frac{1}{4^{2}}\).
  • Simplifying further leads to \(f(4) = 4-\frac{1}{16}\).
  • Finally, this equals \(\frac{63}{16}\).
By evaluating the function, we find that \(f(x)\) produces the output \(\frac{63}{16}\) when \(x = 4\). This systematic approach ensures accuracy in determining the value of the function.
Undefined Functions
Undefined functions occur when an input value does not belong to the function's domain. It is essential to identify these values because attempting to evaluate the function at these points will lead to errors or 'undefined' results. In our example function \(f(x)=x-\frac{1}{x^{2}}\), scenarios where \(x\) is zero or negative cause problems:
  • \(x = 0\): This value is not part of the domain \((0,+\infty)\), thus \(f(x)\) is undefined here since division by zero is invalid.
  • \(x = -1\): Since the domain only allows positive real numbers, \(x = -1\) makes the function undefined.
Properly identifying and avoiding these values prevents mathematical errors and guides you toward correct calculations.
Sequences and Series
Sequences and series refer to ordered lists of numbers and their sums, often appearing in calculus to explore patterns and convergence. While the original exercise focuses on the function \(f(x)=x-\frac{1}{x^{2}}\), sequences and series relate to extending these function evaluations systematically.
Although not directly addressed in the given problem, understanding sequences and series enhances comprehension of how functions behave over intervals. They are especially useful when a function is evaluated repeatedly over a range of values, helping identify trends or repetitive behaviors:
  • Sequences are like lists, such as \(a_1, a_2, a_3, \ldots\)
  • Series are formed by summing sequence terms, for example, \(a_1 + a_2 + a_3 + \ldots\)
These concepts aid in approximating complex functions and are vital in advanced calculus topics.

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Most popular questions from this chapter

In the Fahrenheit temperature scale, water freezes at \(32^{\circ} \mathrm{F}\) and boils at \(212^{\circ} \mathrm{F}\). In the Celsius scale, water freezes at \(0^{\circ} \mathrm{C}\) and boils at \(100^{\circ} \mathrm{C}\). Assuming that the Fahrenheit temperature \(F\) and the Celsius temperature \(C\) are related by a linear equation, find \(F\) in terms of \(C\). Use your equation to find the Fahrenheit temperatures corresponding to \(30^{\circ} \mathrm{C}, 22^{\circ} \mathrm{C},-10^{\circ} \mathrm{C}\), and \(-14^{\circ} \mathrm{C}\), to the nearest degree.

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