Question

What points, if any, are the functions discontinuous? $$ f(x)=\frac{33-x^{2}}{x \pi+3 x-3 \pi-x^{2}} $$

Short answer

The function is discontinuous at \(x = 0\) and \(x = \pi\).

Step-by-step solution

Identify the function type

The given function is a rational function, which is of the form \( \frac{p(x)}{q(x)} \) where both \( p(x) \) and \( q(x) \) are polynomials.

Determine where the denominator is zero

To find discontinuities, we need to determine for which values of \( x \) the denominator \( x\pi + 3x - 3\pi - x^2 = 0 \). These are the potential points of discontinuity.

Solve the quadratic equation

Rewrite the denominator as \(-x^2 + (\pi+3)x - 3\pi\). Set it equal to zero: \[-x^2 + (\pi+3)x - 3\pi = 0\] This is a standard quadratic equation \(ax^2 + bx + c = 0\), where \(a = -1\), \(b = \pi + 3\), and \(c = -3\pi\).

Use the quadratic formula

The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substitute \(a = -1\), \(b = \pi + 3\), and \(c = -3\pi\) into the formula:\[x = \frac{-(\pi + 3) \pm \sqrt{(\pi + 3)^2 - 4(-1)(-3\pi)}}{-2}\]

Simplify the expression under the square root

Calculate \((\pi + 3)^2 - 4(-1)(-3\pi)\):\[(\pi + 3)^2 = \pi^2 + 6\pi + 9\]\[-4(-1)(-3\pi) = 12\pi\]Simplify to get:\[\pi^2 + 6\pi + 9 - 12\pi = \pi^2 - 6\pi + 9\]

Solve for the square root

Since the expression inside the square root is \[\pi^2 - 6\pi + 9 = (\pi - 3)^2\],the square root simplifies to \[\pi - 3\].

Determine the roots

Substitute back into the quadratic formula:\[x = \frac{-(\pi + 3) \pm (\pi - 3)}{-2}\]Calculate both possible roots:1. \[x = \frac{-(\pi + 3) + (\pi - 3)}{-2} = 0\]2. \[x = \frac{-(\pi + 3) - (\pi - 3)}{-2} = \pi\]

Identify points of discontinuity

The function is discontinuous where the denominator is zero. From Step 7, we found the function is discontinuous at \(x = 0\) and \(x = \pi\).

Key concepts

Rational Function

A rational function is a fraction where both the numerator and the denominator are polynomials. For example, in the function \( f(x)=\frac{33-x^{2}}{x \pi+3 x-3 \pi-x^{2}} \),the numerator is \( 33-x^2 \) and the denominator is \( x \pi+3x-3\pi-x^2 \).
Rational functions are an important part of algebra because they can model real-world situations, like rates and densities. They are called "rational" because they involve ratios.Here are some key points about rational functions:

• The domain includes all real numbers except those which make the denominator zero.
• Discontinuities occur at points where the denominator doesn't equal zero.

Understanding rational functions helps in various mathematical applications, particularly in calculus.

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree two. It generally looks like \( ax^2 + bx + c = 0 \).In our example, the denominator of the rational function turns into a quadratic equation: \[-x^2 + (\pi+3)x - 3\pi = 0\].
Quadratic equations can be solved using various methods, such as factoring, completing the square, or applying the quadratic formula.For students, understanding how to recognize and solve quadratic equations is crucial because they often appear in many areasof mathematics and science.
An interesting feature of quadratic equations is their graph, which is a parabola.Parabolas have properties like symmetry and a vertex, which are useful when analyzing the curve.This makes understanding quadratic equations essential not only in solving algebraic problems but also in understanding their geometric representations.

Quadratic Formula

The quadratic formula provides a straightforward way to find the solutions of a quadratic equation.It is defined as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^2 + bx + c = 0 \).
In our example, we used the quadratic formula to find the roots of the equation:\[-x^2 + (\pi+3)x - 3\pi = 0\].To do this, we substituted \( a = -1 \), \( b = \pi + 3 \), and \( c = -3\pi \) into the formula, resulting in:\[ x = \frac{-(\pi + 3) \pm \sqrt{(\pi + 3)^2 - 4(-1)(-3\pi)}}{-2} \].
The quadratic formula is a versatile tool. It can solve any quadratic equation, regardless of whether it can be factored easily or not.As a student, mastering the quadratic formula is pivotal because it simplifies finding the roots of any quadratic equation.

Roots of a Polynomial

Roots of a polynomial are values of \( x \) that make the polynomial equal to zero.In simpler terms, they are where the graph of the polynomial crosses the x-axis.For our function, finding where the denominator is zero helps us identify discontinuities.
In solving the quadratic \(-x^2 + (\pi+3)x - 3\pi = 0\), we found the roots to be \( x = 0 \) and \( x = \pi \).These roots are crucial because they indicate points of discontinuity in the rational function.
Roots give important information about the behavior of polynomials. They help determine interval testing and sign changes,which are useful for sketching graphs and solving inequalities.In calculus, understanding the roots is essential for finding the intercepts and analyzing the function's behavior across different domains.