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Chapter 9: Problem 47

Find the values of \(x\) for which each function is continuous. \(f(x)=\frac{2}{x^{2}+1}\)

Short Answer

Expert verified
The function \(f(x)=\frac{2}{x^{2}+1}\) is continuous for all x 鈭� R (all real numbers), as there are no real values of x that would make the denominator equal to 0.

Step by step solution

01

Determine the Domain of the Function

: First, let's find the domain of the function, which is the set of all x values for which the function is defined. Since our function is a rational function, it will be defined for all x values except where the denominator is equal to 0.
02

Analyze the Denominator

: To ensure that our function is continuous, the denominator of our function should not be equal to 0. Let's verify this: Denominator: \(x^2 + 1\) To see if the denominator is ever equal to 0, we can set up an equation and try to solve for x: \(x^2 + 1 = 0\) From this equation, it's clear that there are no real values of x that would make the denominator equal to 0.
03

Determine the Continuity

: Since we were unable to find any values of x that would make the denominator equal to 0, we can conclude that our function is continuous for all real values of x. The function \(f(x)=\frac{2}{x^{2}+1}\) is continuous for all x 鈭� R (all real numbers).

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Key Concepts

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

Rational Functions
Rational functions are a special type of mathematical function expressed as the ratio of two polynomials. They take the form:
  • \( f(x) = \frac{P(x)}{Q(x)} \)
where \(P(x)\) and \(Q(x)\) are polynomials.
The denominator polynomial \(Q(x)\) must not be zero because division by zero is undefined. This constraint dictates the values that \(x\) can take for the function to remain valid.
Unlike some other functions, rational functions can have breaks or holes at points where \(Q(x) = 0\), leading to discontinuity. This characteristic makes analyzing rational functions an essential topic when studying the continuity of functions.
Domain of a Function
The domain of a function is the complete set of values for which the function is defined. For rational functions, it's necessary to identify where the denominator is zero, as these points must be excluded from the domain. Let鈥檚 consider the example function:
  • \( f(x) = \frac{2}{x^2 + 1} \)
Here, to find the domain, we set the denominator equal to zero:
\(x^2 + 1 = 0\)After solving, we find that no real values of \(x\) satisfy this equation, which means the denominator never becomes zero with real numbers acting as inputs. Consequently, the domain of \(f(x)\) includes all real numbers \(\mathbb{R}\), since the function remains defined across this entire set of inputs.
Real Numbers
Real numbers are the set of numbers that include all the rational and irrational numbers. They can be visualized on a continuous number line with no gaps. Real numbers include:
  • Positive numbers (e.g., 1, 10.5)
  • Negative numbers (e.g., -5, -3.14)
  • Zero (0)
In the context of continuity, a function is continuous over the real numbers if there are no breaks, jumps, or holes when its graph is drawn. For the function
\( f(x) = \frac{2}{x^2 + 1} \),there are no values of \(x\) among the real numbers that would cause a discontinuity (i.e., no values make \(x^2 + 1 = 0\)).
Therefore, when we describe a function as continuous across all real numbers, it means the function's graph remains unbroken over the entire set of real numbers.

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Most popular questions from this chapter

EFFECT OF LUXURY TAX ON CONSUMPTION Government economists of a developing country determined that the purchase of imported perfume is related to a proposed "luxury tax" by the formula \(N(x)=\sqrt{10,000-40 x-0.02 x^{2}} \quad(0 \leq x \leq 200)\) where \(N(x)\) measures the percentage of normal consumption of perfume when a "luxury tax" of \(x \%\) is imposed on it. Find the rate of change of \(N(x)\) for taxes of \(10 \%, 100 \%\), and \(150 \%\).

Find the derivative of each function. \(h(t)=t^{2}(3 t+4)^{3}\)

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Suppose \(F(x)=f\left(x^{2}+1\right)\). Find \(F^{\prime}(1)\) if \(f^{\prime}(2)=3\).

Find \(\frac{d y}{d u^{\prime}} \frac{d u}{d x^{\prime}}\) and \(\frac{d y}{d x}\). \(y=2 u^{2}+1\) and \(u=x^{2}+1\)
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