Determine if Continuous f(x)=x/(x^2+x+8)
The question asks to evaluate if the function f(x) = x/(x^2 + x + 8) is continuous. It implies that you need to investigate the continuity of the given rational function, which is a function where both the numerator and the denominator are polynomials. Continuity, in this context, means that there should be no breaks, jumps, or holes in the graph of the function for all values in its domain. You likely need to examine the behavior of the function around the points where the denominator could be zero, as well as its behavior over the rest of its domain.
$f \left(\right. x \left.\right) = \frac{x}{x^{2} + x + 8}$
Answer
Solution:
Step 1: Identify the domain to check for continuity.
Step 1.1: Equate the denominator of $\frac{x}{x^2 + x + 8}$ to zero to find points of discontinuity.
$x^2 + x + 8 = 0$
Step 1.2: Determine the values of $x$.
Step 1.2.1: Apply the quadratic formula to find the roots.
$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Step 1.2.2: Insert $a = 1$, $b = 1$, and $c = 8$ into the quadratic formula and compute $x$.
$\frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$
Step 1.2.3: Simplify the expression.
Step 1.2.3.1: Simplify the square root in the numerator.
Step 1.2.3.1.1: Recognize that any number raised to the power of one remains the same.
$x = \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$
Step 1.2.3.1.2: Perform the multiplication inside the square root.
Step 1.2.3.1.2.1: Multiply $-4$ by $1$.
$x = \frac{-1 \pm \sqrt{1 - 4 \cdot 8}}{2 \cdot 1}$
Step 1.2.3.1.2.2: Multiply $-4$ by $8$.
$x = \frac{-1 \pm \sqrt{1 - 32}}{2 \cdot 1}$
Step 1.2.3.1.3: Subtract $32$ from $1$.
$x = \frac{-1 \pm \sqrt{-31}}{2 \cdot 1}$
Step 1.2.3.1.4: Express $-31$ as $-1 \cdot 31$.
$x = \frac{-1 \pm \sqrt{-1 \cdot 31}}{2 \cdot 1}$
Step 1.2.3.1.5: Rewrite $\sqrt{-1 \cdot 31}$ as $\sqrt{-1} \cdot \sqrt{31}$.
$x = \frac{-1 \pm \sqrt{-1} \cdot \sqrt{31}}{2 \cdot 1}$
Step 1.2.3.1.6: Replace $\sqrt{-1}$ with $i$.
$x = \frac{-1 \pm i \sqrt{31}}{2 \cdot 1}$
Step 1.2.3.2: Simplify the denominator.
$x = \frac{-1 \pm i \sqrt{31}}{2}$
Step 1.2.4: Combine both solutions for the final answer.
$x = -\frac{1 - i \sqrt{31}}{2}, -\frac{1 + i \sqrt{31}}{2}$
Step 1.3: Since the solutions are not real numbers, the domain is all real numbers.
Interval Notation: $(-\infty, \infty)$ Set-Builder Notation: $\{x | x \in \mathbb{R}\}$
Step 2: Given that the domain includes all real numbers, $\frac{x}{x^2 + x + 8}$ is continuous for all real numbers.
Step 3: There is no further action required as the function is continuous over its entire domain.
Knowledge Notes:
To determine the continuity of a function, one must first identify its domain, which is the set of all input values for which the function is defined. For rational functions, such as $\frac{x}{x^2 + x + 8}$, the domain is all real numbers except for those that make the denominator equal to zero.
The quadratic formula is used to find the roots of a quadratic equation $ax^2 + bx + c = 0$. The formula is given by $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. If the discriminant ($b^2 - 4ac$) is negative, the roots are complex numbers and do not affect the domain of a function defined on the real numbers.
Complex numbers are numbers in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit, defined as $\sqrt{-1}$. Since complex roots do not intersect the x-axis, they do not create points of discontinuity in the real-valued function.
Interval notation is a way of writing subsets of the real number line. An interval that includes all real numbers is written as $(-\infty, \infty)$. Set-builder notation is another way to describe sets, using a rule to define the elements of the set. For the set of all real numbers, it is written as $\{x | x \in \mathbb{R}\}$, where $\mathbb{R}$ represents the set of all real numbers.
A function is continuous if it does not have any breaks, holes, or jumps in its graph. If a function is defined and continuous over its entire domain, then it can be said to be continuous for all real numbers.
