Determine if Continuous f(x)=x/(x^2+x+5)

Solution:

Determine the Continuity of $f(x) = \frac{x}{x^2 + x + 5}$

Step:1

Identify the domain to check the continuity of the function.

Step:1.1

Equating the denominator of $\frac{x}{x^{2} + x + 5}$ to zero to find the points of discontinuity: $x^{2} + x + 5 = 0$

Step:1.2

Find the values of $x$.

Step:1.2.1

Apply the quadratic formula: $\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

Step:1.2.2

Insert $a = 1$, $b = 1$, and $c = 5$ into the quadratic formula and solve for $x$: $\frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step:1.2.3

Simplify the expression.

Step:1.2.3.1

Simplify the numerator.

Step:1.2.3.1.1

Any number raised to the power of one remains the same: $x = \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step:1.2.3.1.2

Perform the multiplication of $-4 \cdot 1 \cdot 5$.

Step:1.2.3.1.2.1

Multiply $-4$ by $1$: $x = \frac{-1 \pm \sqrt{1 - 4 \cdot 5}}{2 \cdot 1}$

Step:1.2.3.1.2.2

Multiply $-4$ by $5$: $x = \frac{-1 \pm \sqrt{1 - 20}}{2 \cdot 1}$

Step:1.2.3.1.3

Subtract $20$ from $1$: $x = \frac{-1 \pm \sqrt{-19}}{2 \cdot 1}$

Step:1.2.3.1.4

Express $-19$ as $-1 \cdot 19$: $x = \frac{-1 \pm \sqrt{-1 \cdot 19}}{2 \cdot 1}$

Step:1.2.3.1.5

Represent $\sqrt{-1 \cdot 19}$ as $\sqrt{-1} \cdot \sqrt{19}$: $x = \frac{-1 \pm \sqrt{-1} \cdot \sqrt{19}}{2 \cdot 1}$

Step:1.2.3.1.6

Replace $\sqrt{-1}$ with $i$: $x = \frac{-1 \pm i \sqrt{19}}{2}$

Step:1.2.3.2

Multiply $2$ by $1$: $x = \frac{-1 \pm i \sqrt{19}}{2}$

Step:1.2.4

Combine both solutions for the final answer: $x = -\frac{1 - i \sqrt{19}}{2}, -\frac{1 + i \sqrt{19}}{2}$

Step:1.3

The domain consists of 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, the function $\frac{x}{x^{2} + x + 5}$ is continuous for all real numbers.

Step:3

No further steps are required as the function is continuous across its 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, like the one given, discontinuities can occur where the denominator is zero, as division by zero is undefined.

The quadratic formula is used to solve quadratic equations of the form $ax^2 + bx + c = 0$. The formula is given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. If the discriminant ($b^2 - 4ac$) is negative, the solutions are complex numbers, which do not affect the real domain of the function.

Complex numbers are of the form $a + bi$, where $i$ is the imaginary unit, defined as $\sqrt{-1}$. In this case, since the solutions are complex, they do not restrict the domain of the function, which remains all real numbers.

Continuity of a function at a point means that there is no interruption in the graph of the function at that point. A function is continuous over an interval if it is continuous at every point in that interval. If a function is defined and continuous over its entire domain, it is simply called continuous.