Determine if Continuous f(x)=x/(x^2+x+3)

Solution:

Step 1: Identify the domain to verify continuity of the function.

• Step 1.1: Equate the denominator of $\frac{x}{x^2+x+3}$ to zero to find points of discontinuity.

  • $x^2+x+3=0$

• Step 1.2: Determine the values of $x$ that satisfy the equation.

  • 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 $c=3$ into the formula and calculate $x$.

    • $\frac{-1\pm \sqrt{1^2-4\cdot 1\cdot 3}}{2\cdot 1}$

  • 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 unchanged.

        • $x=\frac{-1\pm \sqrt{1-4\cdot 1\cdot 3}}{2\cdot 1}$

      • Step 1.2.3.1.2: Perform the multiplication inside the square root.

        • Step 1.2.3.1.2.1: Multiply $-4$ by $1$.

          • $x=\frac{-1\pm \sqrt{1-4\cdot 3}}{2\cdot 1}$

        • Step 1.2.3.1.2.2: Multiply $-4$ by $3$.

          • $x=\frac{-1\pm \sqrt{1-12}}{2\cdot 1}$

      • Step 1.2.3.1.3: Subtract $12$ from $1$.

        • $x=\frac{-1\pm \sqrt{-11}}{2\cdot 1}$

      • Step 1.2.3.1.4: Express $-11$ as $-1\cdot 11$.

        • $x=\frac{-1\pm \sqrt{-1\cdot 11}}{2\cdot 1}$

      • Step 1.2.3.1.5: Rewrite $\sqrt{-1\cdot 11}$ as $\sqrt{-1}\cdot \sqrt{11}$.

        • $x=\frac{-1\pm \sqrt{-1}\cdot \sqrt{11}}{2\cdot 1}$

      • Step 1.2.3.1.6: Replace $\sqrt{-1}$ with $i$.

        • $x=\frac{-1\pm i\sqrt{11}}{2\cdot 1}$

    • Step 1.2.3.2: Multiply $2$ by $1$.

      • $x=\frac{-1\pm i\sqrt{11}}{2}$

  • Step 1.2.4: Combine both solutions to get the final answer.

    • $x=-\frac{1-i\sqrt{11}}{2},-\frac{1+i\sqrt{11}}{2}$

• Step 1.3: Since the solutions are not real numbers, the domain is all real numbers.

  • Interval Notation: $(-\mathrm{\infty },\mathrm{\infty })$
  • Set-Builder Notation: $\{x|x\in \mathbb{R}\}$

Step 2: With the domain being all real numbers, the function $\frac{x}{x^2+x+3}$ is continuous for all real numbers.

Step 3: There are no points of discontinuity, and the function is continuous everywhere on its domain.

Knowledge Notes:

The problem involves determining the continuity of the function $f(x)=\frac{x}{x^2+x+3}$. To do this, we need to find the domain of the function, which is the set of all input values (x-values) for which the function is defined. A function is continuous at a point if it does not have any breaks, holes, or jumps at that point. For rational functions, like the one given, discontinuities can occur where the denominator is zero, since division by zero is undefined.

The steps of the solution involve:

1. Setting the denominator equal to zero and solving for $x$ to find potential points of discontinuity.

2. Using the quadratic formula to solve for $x$ when the denominator is a quadratic expression.

3. Simplifying the resulting expression to determine if there are any real solutions.

4. Concluding that if there are no real solutions to the denominator being zero, the function's domain is all real numbers, and thus the function is continuous everywhere on its domain.

Relevant knowledge points include:

• The definition of a continuous function.

• The domain of a function, which is the set of all possible input values.

• How to use the quadratic formula: $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a$, $b$, and $c$ are coefficients of the quadratic equation $a{x}^2+bx+c=0$.

• Complex numbers and the imaginary unit $i$, where $i^2=-1$.

• Interval notation and set-builder notation for expressing domains of functions.