Problem 44 - Algebra and Trigonometry with Analytic Geometry

Question

Simplify \(f(x)\), and sketch the graph of \(f\). $$ 4 f(x)=\frac{\left(x^{2}+x\right)(2 x-1)}{\left(x^{2}-3 x+2\right)(2 x-1)} $$

Step-by-Step Solution

Answer

Simplified, \(f(x) = \frac{x(x+1)}{(x-1)(x-2)}\). It has discontinuities at \(x=1, 2, \frac{1}{2}\).

Step 1: Identify Common Factors

First, observe the expression for potential cancellation. In the given expression: \( \frac{(x^2+x)(2x-1)}{(x^2-3x+2)(2x-1)} \), there is a common factor \((2x-1)\) in both the numerator and the denominator.

Step 2: Cancel Out Common Factors

Since \((2x-1)\) is a common factor in both the numerator and the denominator, we can cancel it out, simplifying the expression to \( \frac{x^2+x}{x^2-3x+2} \). This is valid for \(x eq \frac{1}{2}\) because dividing by zero is undefined.

Step 3: Factor the Denominator

Factor the denominator \(x^2-3x+2\). This expression can be factored into \((x-1)(x-2)\). Therefore, the simplified expression becomes \( \frac{x^2+x}{(x-1)(x-2)} \).

Step 4: Express the Numerator in Factorized Form

Factor the numerator \(x^2+x\). This can be factored as \(x(x+1)\). Therefore, the simplified expression is now \( \frac{x(x+1)}{(x-1)(x-2)} \).

Step 5: Determine Points of Discontinuity

The original function has discontinuity where any factor of the denominator is zero. The simplified form suggests discontinuity at \(x=1\) and \(x=2\). The cancelled factor \((2x - 1)\) indicates a removable discontinuity at \(x=\frac{1}{2}\).

Step 6: Sketch the Graph of \(f(x)\)

Plot the simplified form of \(f(x) = \frac{x(x+1)}{(x-1)(x-2)}\). Note the vertical asymptotes at \(x=1\) and \(x=2\). The function is undefined at \(x=\frac{1}{2}\), creating a removable discontinuity (hole) here. Intercepts can be calculated by setting the numerator equal to zero. The x-intercepts occur at \(x=0\) and \(x=-1\).

Key Concepts

Rational Functions

In algebra, a rational function is simply a fraction where both the numerator and the denominator are polynomials. Both parts can be of any degree as long as the denominator is not zero. For our function, we initially have \[ f(x) = \frac{(x^2 + x)(2x-1)}{(x^2 - 3x + 2)(2x-1)} \].
It's important to locate common factors between the numerator and the denominator to simplify the function, a key step in handling rational functions. This simplifying process helps reveal the essential characteristics of the function's behavior.

• Cancellation: Reduces the overall degree and complexity.
• Focuses only on points where the function might otherwise be undefined due to zero-denominator issues.

By canceling the shared factor \((2x-1)\), we make the expression much easier to work with while also identifying potential points of discontinuity when previously shared factors come into play.

Discontinuities

Points of discontinuities in functions, especially rational functions, occur primarily when the denominator is zero, as a function cannot have a zero value in the denominator. In our problem, discontinuities are rooted in the denominator \((x^2 - 3x + 2)(2x - 1)\) since these terms cannot equal zero. Discontinuities appear as vertical asymptotes or removable discontinuities.
To find a discontinuity:

• Set the denominator equal to zero and solve. For example, \((x-1)(x-2)\) suggests discontinuities at \(x=1\) and \(x=2\).
• A removable discontinuity occurs when a factor leading to zero in the denominator is canceled out in the numerator, such as \((2x-1)\) at \(x=\frac{1}{2}\).

Visible discontinuities, like holes, occur at these removable discontinuities and are generally plotted as open circles on graphs. With this information, we can then consider these areas in graph sketching.

Graph Sketching

Graph sketching of rational functions involves using identified discontinuities and other critical characteristics such as intercepts. After simplifying the function, the challenging part is visualizing what it looks like graphically. Here's a step-by-step approach to sketch the graph of \[ f(x) = \frac{x(x+1)}{(x-1)(x-2)} \].
The steps include:

• Identify the x-intercepts by setting the numerator equal to zero: \(x = 0\) and \(x = -1\).
• Determine the vertical asymptotes, which occur at the points where each factor of the denominator is zero: \(x = 1\) and \(x = 2\).
• Locate removable discontinuities, plotted as holes, at points where factors have been cancelled: hole at \(x=\frac{1}{2}\).

Include horizontal evaluations for end behavior. By examining these aspects, you strategically place different characteristics that visually depict the behavior of the function across its domain. Such sketching helps in understanding how rational functions handle discontinuities and complex polynomial behaviors. Also, it is important to remember that these asymptotes guide the path which the functions approach but never reach.