Problem 12 - 5 Steps to A 5: AP Calculus AB 2021

Question

Sketch the graphs of the following functions indicating any relative and absolute extrema, points of inflection, intervals on which the function is increasing, decreasing, concave upward, or concave downward. $$ f(x)=\frac{x+4}{x-4} $$

Step-by-Step Solution

Answer

**Question**: Sketch the graph of the function $f(x) = \frac{x+4}{x-4}$. Describe its key features, such as relative/absolute extrema, points of inflection, intervals in which the function is increasing/decreasing, and intervals of concavity. **Answer**: The graph of the function $f(x) = \frac{x+4}{x-4}$ has the following key features: 1. The function is increasing on the interval $(-\infty, 4)$ and decreasing on the interval $(4, \infty)$. 2. There is a vertical asymptote at x = 4. 3. There is a point of inflection at x = 4, but it is not on the graph due to the asymptote. 4. The function is concave downward on the interval $(-\infty, 4)$ and concave upward on the interval $(4, \infty)$.

Step 1: Find the first derivative

To find the intervals where the function is increasing or decreasing, we need to find the first derivative of the function. Using the quotient rule, we get: $$ f'(x) = \frac{(x-4)(1) - (x+4)(1)}{(x-4)^2} = \frac{-8}{(x-4)^2} $$

Step 2: Find the critical points

To find the critical points, we must determine the values of x where the first derivative is equal to zero or undefined. In this case, the first derivative is never equal to zero. However, it is undefined when the denominator is equal to zero: $$ (x-4)^2 = 0 \Rightarrow x = 4 $$ Thus, x = 4 is a critical point.

Step 3: Determine increasing/decreasing intervals

To find the intervals where the function is increasing or decreasing, we can use the first derivative test. We'll analyze the signs of the first derivative at intervals separated by the critical point x = 4: If x > 4, the first derivative is negative \((f'(x)=-8/(x-4)^2<0)\). Therefore, the function is decreasing for \(x > 4\). If x < 4, the first derivative is positive \((f'(x)=-8/(x-4)^2>0)\). Therefore, the function is increasing for \(x < 4\).

Step 4: Find the second derivative

To find the intervals where the function is concave upward or downward and to find the points of inflection, we need to find the second derivative of the function. We will also use this to find any relative extrema. Taking the derivative of \(f'(x)\), we get: $$ f''(x) = \frac{16(x-4)}{(x-4)^4} $$

Step 5: Determine concavity and points of inflection

To find the intervals where the function is concave upward or downward and to find the points of inflection, we will analyze the signs of the second derivative. f''(x) is undefined when x = 4, which corresponds to a vertical asymptote at x = 4. If x > 4, then the second derivative is positive \((f''(x)=16(x-4)/(x-4)^4>0)\). Therefore, the function is concave upward for \(x > 4\). If x < 4, then the second derivative is negative \((f''(x)=16(x-4)/(x-4)^4<0)\). Therefore, the function is concave downward for \(x < 4\). Since the concavity changes at x = 4, there's a point of inflection at x = 4. However, since this point corresponds to a vertical asymptote, it does not lie on the graph of the function.

Step 6: Sketch the graph

Using the information gathered, sketch the graph of the function: 1. The function is increasing on \((-\infty, 4)\) and decreasing on \((4, \infty)\). 2. There is an asymptote at x = 4. 3. There is a point of inflection at x = 4, but it is not on the graph due to the asymptote. 4. The function is concave downward on \((-\infty, 4)\) and concave upward on \((4, \infty)\).

Key Concepts

Graphing Rational Functions

Understanding how to graph rational functions is a critical skill in AP Calculus AB. A rational function, like \(f(x)=\frac{x+4}{x-4}\), is characterized by the division of two polynomials.

The first step in graphing a rational function involves identifying any asymptotes, which are lines that the graph approaches but never touches. Vertical asymptotes occur where the denominator is zero and the function is undefined. In our example, \(x-4=0\) implies a vertical asymptote at \(x=4\).

Horizontal asymptotes depend on the degrees of the polynomials in the numerator and the denominator. Given that the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. However, since the degrees are the same and the leading coefficients are 1 and 1, the horizontal asymptote for our function is at \(y = 1\).

• Plot the asymptotes on the graph to create a framework for the sketch.
• Next, determine the x-intercept (if any) by setting the numerator equal to zero and solving for x, which yields the x-intercept as \(x=-4\).
• Consider the end behavior of the function, often indicated by the asymptotes and intercepts.

Finally, include points on both sides of the vertical asymptote to understand how the function behaves and then sketch the function along these guidelines.

First Derivative Test

The first derivative test is a useful technique for analyzing the intervals on which a function is increasing or decreasing. By taking the derivative of the function, \(f'(x)\), and determining its sign, we can identify the behavior of the function.

In our example, the derivative \(f'(x) = \frac{-8}{(x-4)^2}\) informs us that the function has a critical point at \(x=4\) because the derivative is undefined there.

• For \(x < 4\), \(f'(x) > 0\), which means our function is increasing on this interval.
• For \(x > 4\), \(f'(x) < 0\), indicating a decreasing behavior for the function.

To use the first derivative test effectively, create a sign chart that displays the intervals divided by the critical points and shows the sign of the derivative in each interval. This will help you discover where the function's graph is going upwards or downwards.

Concavity and Points of Inflection

Concavity refers to the direction of the curve of a function, and the points of inflection are where the concavity changes from up to down or vice versa. To analyze concavity, we use the second derivative \(f''(x)\).

For the function \(f(x)=\frac{x+4}{x-4}\), the second derivative is \(f''(x) = \frac{16(x-4)}{(x-4)^4}\). The sign of this derivative will tell us about the concavity:

• If \( f''(x) > 0\), the function is concave upward.
• If \( f''(x) < 0\), the function is concave downward.

Our function does not change concavity because the second derivative has the same sign on both sides of \(x=4\). However, since \(x=4\) is a vertical asymptote and represents a point at which the function is not defined, it would be a potential point of inflection if it were not for the discontinuity at that point.

Critical Points and Extrema

Critical points are where a function's derivative is either zero or undefined and can indicate possible locations for relative extrema, which are the highest or lowest points in a certain interval of the graph.

To find critical points, like we did for our function \(f(x)=\frac{x+4}{x-4}\), we look for where \(f'(x)=0\) or is undefined. The critical point of our function is \(x=4\), but since this corresponds to a vertical asymptote, it does not result in a relative extremum on the graph.

To identify absolute extrema, which are the highest or lowest points on the entire graph, we must examine the end behavior and any local extrema. Since the function has no local maxima or minima and approaches different limits as \(x\) approaches positive and negative infinity, the function has no absolute extrema.

• Consider the critical points along with the first and second derivatives.
• Analyze the behavior near the critical points to determine if they are relative maxima, minima, or neither.

For rational functions like ours, understanding asymptotic behavior is crucial in identifying the absence of local and absolute extrema.