Problem 38 - Essential Calculus Early Transcendentals

Question

(a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts ( d ) to sketch the graph of \(f .\) \(f(x)=\frac{x^{2}-4}{x^{2}+4}\)

Step-by-Step Solution

Answer

Horizontal asymptote at \(y=1\), local minimum at \(x=0, y=-1\), concave up on \((-\infty, -2)\) and \((2, \infty)\), concave down on \((-2, 2)\).

Step 1: Identify Vertical Asymptotes

For vertical asymptotes, set the denominator equal to zero. The function is \(f(x) = \frac{x^{2} - 4}{x^{2} + 4}\). Since \(x^{2} + 4\) is always positive (never zero), there are no vertical asymptotes.

Step 2: Identify Horizontal Asymptotes

Divide the coefficients of the highest powers in the numerator and denominator. Since both numerator and denominator have the same degree (2), the horizontal asymptote is\(y = \frac{1}{1} = 1\).

Step 3: Find Critical Points

Find the derivative, \(f'(x) = \frac{(2x)(x^{2}+4) - (2x)(x^{2}-4)}{(x^{2}+4)^2}\). Simplify it to \(f'(x) = \frac{16x}{(x^{2}+4)^2}\). Set \(f'(x) = 0\) to find the critical points: \(16x = 0\) gives \(x = 0\).

Step 4: Determine Intervals of Increase/Decrease

Test the intervals around the critical point \(x = 0\). Choose \(x = 1\) and \(x = -1\) to test the sign of \(f'(x)\). We find \(f'(1) > 0\) and \(f'(-1) < 0\), indicating the function is decreasing on \((-\infty, 0)\) and increasing on \((0, \infty)\).

Step 5: Assess Local Extrema

Use the results from Step 4: since \(f(x)\) transitions from decreasing to increasing at \(x = 0\), there is a local minimum at \(x = 0\). \(f(0) = \frac{-4}{4} = -1\), so the local minimum is \((-1)\).

Step 6: Find Concavity and Inflection Points

Compute the second derivative \(f''(x)\). Simplifying the initially complex derivative yields \(f''(x) = \frac{48(x^2-4)}{(x^2+4)^3}\). Set \(f''(x) = 0\) to check potential inflection points: \(x = \pm2\). Test intervals to confirm concavity.

Step 7: Analyze Concavity

Evaluate \(f''(x)\) around \(x = -2\) and \(x = 2\). For \((-\infty, -2)\), \(f''(x) > 0\); for \((-2, 2)\), \(f''(x) < 0\); and for \((2, \infty)\), \(f''(x) > 0\). Therefore, the graph is concave up on \((-\infty, -2)\) and \((2, \infty)\), and concave down on \((-2, 2)\).

Step 8: Sketch the Graph

Using the asymptotes, increasing/decreasing intervals, and concavity from prior steps, sketch the function. Note the horizontal asymptote at \(y=1\), local minimum at \((0, -1)\), intervals of increase/decrease, and concavity for shaping the graph.

Key Concepts

Asymptotes

Understanding asymptotes is essential when analyzing the behavior of a function. Asymptotes are lines that a graph approaches but never touches. They help us predict the behavior of graphs at extreme values.

• Vertical Asymptotes: These occur when the denominator of a function approaches zero. However, for the function \(f(x) = \frac{x^{2} - 4}{x^{2} + 4}\), the denominator is always positive and never zero. Thus, there are no vertical asymptotes.
• Horizontal Asymptotes: These are determined by comparing the degrees of the numerator and denominator. Here, both have degree 2, so the horizontal asymptote is \(y = \frac{1}{1} = 1\). This means that as \(x\) approaches infinity, \(f(x)\) approaches 1.

Recognizing these asymptotes helps us understand how the function behaves at its extremes.

Critical Points

Critical points are where the function stops increasing or decreasing. They occur where the first derivative is zero or undefined. Finding these points helps us locate points of interest such as minimums or maximums.
To find critical points for \(f(x)\), first calculate the derivative, \(f'(x) = \frac{16x}{(x^{2}+4)^2}\). Setting \(f'(x) = 0\) gives us \(x = 0\) as the critical point. This critical value indicates where the function might change direction.

• Check intervals around this critical point by selecting test values like \(x = 1\) and \(x = -1\) to determine if \(f(x)\) is increasing or decreasing.
• Results show \(f(x)\) decreases on \((-fi, 0)\) and increases on \((0, \infty)\).

The critical point, \(x = 0\), turns out to be a location of a local minimum as the function switches from decreasing to increasing.

Concavity

The concavity of a function tells us how it curves and helps identify potential turning points. We determine concavity by finding the second derivative.
After obtaining ©\(f''(x) = \frac{48(x^2-4)}{(x^2+4)^3}\), examine the sign of \(f''(x)\) across different intervals.

• For \(x < -2\), \(f''(x)\) is positive, indicating the graph is concave up.
• For \(-2 < x < 2\), \(f''(x)\) becomes negative, showing the graph is concave down.
• Lastly, for \(x > 2\), \(f''(x)\) is positive again, meaning the graph is concave up.

Understanding these intervals of concavity is crucial for sketching the overall shape of the graph. It tells us when the graph changes from a "cup" shape to a "cap" shape, or vice versa.

Inflection Points

Inflection points occur where the function changes concavity. These points signify a change in the direction of curvature on a graph. To find inflection points for \(f(x)\), solve the equation \(f''(x) = 0\). This gives potential inflection points at \(x = \pm2\). Test intervals around these points to confirm changes in concavity.

• At \(.x = -2\) and \(x = 2\), the sign of the second derivative changes, indicating these are indeed inflection points.

Recognizing inflection points provides crucial data when sketching or interpreting the graph of a function, as they indicate critical changes in the graph's pathway.