Problem 88 - Essential Calculus

Question

In Exercises 87 and \(88,\) (a) use a graphing utility to graph \(f\) and \(g\) in the same viewing window, (b) verify algebraically that \(f\) and \(g\) represent the same function, and (c) zoom out sufficiently far so that the graph appears as a line. What equation does this line appear to have? (Note that the points at which the function is not continuous are not readily seen when you zoom out.) $$ \begin{array}{l} f(x)=-\frac{x^{3}-2 x^{2}+2}{2 x^{2}} \\ g(x)=-\frac{1}{2} x+1-\frac{1}{x^{2}} \end{array} $$

Step-by-Step Solution

Answer

On verifying algebraically, it is confirmed that both \(f(x)\) and \(g(x)\) represent the same function \(f(x) = g(x) = -\frac{1}{2}x + 1 - \frac{1}{x^2}\). Upon zooming out far enough on the graph, it should appear as a line, the equation of which can be determined from the graph itself.

Step 1: Graph the Functions

Use a graphing utility to graph both \(f(x) = -\frac{x^{3}-2 x^{2}+2}{2 x^{2}} \) and \(g(x) = -\frac{1}{2} x+1-\frac{1}{x^{2}} \). Plot both graphs in the same viewing window to visualize their similarity and differences, if any.

Step 2: Algebraic Verification

Verify algebraically whether the two equations \(f(x)\) and \(g(x)\) depict the same mathematical function. To do this, rewrite the function \(f(x)\) in a simplified form and then compare with \(g(x)\). Let's start by dividing each term in the numerator of \(f(x)\) by \(2x^2\) to get, \(-\frac{1}{2}x + 1 - \frac{1}{x^2}\). Now we can observe that both functions are identical, hence, they are the same function.

Step 3: Determination of Line Equation

Zoom out the graph to a level where it appears to be a straight line. The equation of this line can be found based on the shape and properties of the line visible in the graph. This line will represent the behavior of this function over a larger domain and range.

Key Concepts

Algebraic Verification

Algebraic verification is a critical process in mathematics where we prove that two algebraic expressions represent the same function. In the context of graphing functions, it ensures that the visual resemblance seen on graphs corresponds to mathematical equivalence. To verify expressions algebraically, one might need to simplify, factor, or manipulate the expressions by applying algebraic rules.

For instance, with functions expressed as rational functions like the exercise's given functions, we carefully analyze each term. As demonstrated in the step by step solution, dividing each term of the numerator by the polynomial in the denominator transforms the complex fraction into a simpler form. Once simplified, finding that both expressions reduce to \( -\frac{1}{2}x + 1 - \frac{1}{x^2} \) indicates they are indeed the same function algebraically, confirming what the graph suggests visually.

Rational Functions

Rational functions, such as the ones given in the exercise \( f(x) \) and \( g(x) \) are quotients of two polynomials. They have unique features, including asymptotes, discontinuities, and variable domains. These functions are particularly interesting when graphing because their behavior can illustrate complex interactions between the numerator and the denominator.

For example, a term like \( \frac{1}{x^2} \) in the function \( g(x) \) can produce a vertical asymptote and a discontinuity at \( x=0 \) due to division by zero being undefined. Graphical representations are invaluable for visualizing these aspects; however, understanding the algebraic structure leads to predicting these features without a graph. Simplifying the given \( f(x) \) resulted in the same structure as \( g(x) \)—indicative of shared properties between the two.

Continuity of Functions

Continuity of functions is a foundational concept which deals with how smoothly a function's graph is drawn without lifting the pencil. For a function to be continuous at a point, it must meet three conditions: the function must be defined at that point, it must approach a limit as it nears the point, and the limit must equal the function's value at that point.

In rational functions, discontinuities often occur at points where the denominator equals zero. The exercise exemplifies this with a discontinuity at \( x=0 \) for the function \( g(x) \) due to the term \( -\frac{1}{x^2} \). When zooming out on a graph, fine details such as discontinuities can become less visible, giving the misconception of a continuous function. Therefore, understanding the algebra behind the function is essential to correctly identifying points of continuity and discontinuity.