Problem 20 - A Graphical Approach to Precalculus with Limits

Question

Let \(f(x)=x^{2}+3 x\) and \(g(x)=2 x-1 .\) Perform the composition or operation indicated. $$\left(\frac{f}{g}\right)\left(\frac{1}{2}\right)$$

Step-by-Step Solution

Answer

The expression is undefined because \(g\left(\frac{1}{2}\right) = 0\) leads to division by zero.

Step 1: Understand the Function Composition

The expression \(\left(\frac{f}{g}\right)\left(\frac{1}{2}\right)\) means that we are to evaluate the function \(\frac{f(x)}{g(x)}\) at \(x = \frac{1}{2}\). This involves first determining \(f\left(\frac{1}{2}\right)\) and \(g\left(\frac{1}{2}\right)\), and then dividing these two results.

Step 2: Calculate \(f(x)\) at \(x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into \(f(x) = x^2 + 3x\). \[f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right)\]\[= \frac{1}{4} + \frac{3}{2}\]\[= \frac{1}{4} + \frac{6}{4}\]\[= \frac{7}{4}\]

Step 3: Calculate \(g(x)\) at \(x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into \(g(x) = 2x - 1\).\[g\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right) - 1\]\[= 1 - 1\]\[= 0\]

Step 4: Evaluate \(\frac{f}{g}\) at \(x = \frac{1}{2}\)

Using the results from Steps 2 and 3, evaluate:\[\frac{f\left(\frac{1}{2}\right)}{g\left(\frac{1}{2}\right)} = \frac{\frac{7}{4}}{0}\]Since division by zero is undefined, this expression does not have a value.

Step 5: Conclusion

The expression \(\left(\frac{f}{g}\right)\left(\frac{1}{2}\right)\) is undefined because the denominator \(g\left(\frac{1}{2}\right)\) results in zero.

Key Concepts

Quadratic Functions

Quadratic functions are mathematical expressions where the highest degree of the variable is two. These functions take the form \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. The graph of a quadratic function is a curve called a parabola. Parabolas can open upwards or downwards depending on the sign of \(a\). When \(a > 0\), the parabola opens upwards, creating a U-shape. Conversely, when \(a < 0\), it opens downwards.

• The vertex of the parabola is the highest or lowest point on the graph, depending on its orientation.
• Quadratics often have one or two x-intercepts, which are the points where the graph crosses the x-axis.
• The y-intercept is the value of the function when \(x = 0\).

In the given problem, the quadratic function is \(f(x) = x^2 + 3x\). Here, the coefficients are \(a=1\), \(b=3\), and \(c=0\), so the parabola opens upwards, and its vertex can be found at \(x = -\frac{b}{2a}\). In function composition, understanding the graph can help visualize how changes in \(x\) affect \(f(x)\).

Rational Functions

Rational functions are ratios of two polynomials, expressed as \(R(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomial expressions and \(Q(x) e 0\). They are defined everywhere except where the denominator \(Q(x)\) equals zero, as this causes division by zero, making the function undefined at those points.

• Asymptotes are important features of rational functions, often appearing as vertical, horizontal, or oblique lines.
• Vertical asymptotes occur at the x-values where the denominator is zero.
• Horizontal or slant asymptotes are determined by the degrees of the polynomials in the numerator and denominator.

In our example, \(g(x) = 2x - 1\) is the linear denominator of the rational function \(\frac{f(x)}{g(x)}\). Evaluating \(g(x)\) when \(x = \frac{1}{2}\) gives zero, which demonstrates a key aspect of rational functions: the need to check the denominator to avoid undefined expressions.

Undefined Expressions

Undefined expressions are terms in mathematical functions that do not have a finite value due to issues such as division by zero. In the context of rational functions, we avoid undefined expressions by ensuring the denominator does not equal zero.

• When a denominator is zero, the rational function becomes undefined at that specific point. This results in a vertical asymptote.
• An expression like \(\frac{a}{0}\) is deemed undefined because it doesn't produce a meaningful or real number. Mathematical operations with undefined values are not possible.

In the problem we explored, \(\frac{f}{g}(\frac{1}{2})\) is undefined since \(g(\frac{1}{2}) = 0\). Whenever evaluating rational functions, particularly during composition, always ascertain that the denominator remains non-zero. This precaution helps avoid incorrect conclusions and highlights the importance of recognizing limitations within function domains.