Problem 10 - Calculus Single Variable

Question

Follow the outline given in this section to give a careful sketch of the graph of each of the functions in Exercises \(1-24\). Your sketch should exhibit, and have labeled, all of the following: a) local and global extrema, b) inflection points, c) intervals on which function is increasing or decreasing, d) intervals on which function is concave up or concave down, e) horizontal and vertical asymptotes. $$ f(x)=x /(x-4)^{2} $$

Step-by-Step Solution

Answer

Sketch has a minimum at \(x=-4\), is undefined at \(x=4\), increases on \((-4,4)\), decreases on \((-\infty,-4)\cup(4,\infty)\), and has asymptotes at \(x=4\) (vertical) and \(y=0\) (horizontal).

Step 1: Find critical points

Differentiate the function to find critical points where the first derivative is zero or undefined. The derivative of the function \( f(x) = \frac{x}{(x-4)^2} \) using the quotient rule is \( f'(x) = \frac{(x-4)^2 - 2x(x-4)}{(x-4)^4} = \frac{(x^2 - 8x + 16 - 2x^2 + 8x)}{(x-4)^4} = \frac{-x^2 + 16}{(x-4)^4} \). Set \( f'(x) = 0 \) to find critical points: \( -x^2 + 16 = 0 \), which gives \( x = \pm 4 \). The derivative is also undefined at \( x = 4 \). Hence, critical points are \( x = -4, 4 \).

Step 2: Identify local and global extrema

Evaluate \( f(x) \) at identified critical points to check for local extrema: \( f(-4) = \frac{-4}{64} = -\frac{1}{16} \). Since \( f(x) \) is undefined at \( x = 4 \), check both sides around \( x = 4 \): Near \( 4^- \), \( f(x) \rightarrow -\infty \); Near \( 4^+ \), \( f(x) \rightarrow +\infty \). Thus, there is no global extremum at \( x = 4 \), and a global minimum at \( x = -4 \).

Step 3: Second derivative for concavity and inflection points

Compute the second derivative from \( f'(x) \): \( f''(x) = \frac{d}{dx} \left(\frac{-x^2 + 16}{(x-4)^4}\right) \). Applying the quotient rule yields \( f''(x) = \frac{2x(x-4)^4 - 4(x-4)^3(-x^2 + 16)}{(x-4)^8} \). Simplify further to find inflection points by setting \( f''(x) = 0 \) or identifying where \( f''(x) \) changes sign.

Step 4: Determine intervals of increase and decrease

Use the first derivative \( f'(x) = \frac{-x^2 + 16}{(x-4)^4} \). \( f'(x) > 0 \) implies the function is increasing; \( f'(x) < 0 \) implies decreasing. Solve \( -x^2 + 16 > 0 \) giving \( x \in (-4, 4) \) as the increasing interval, while \( x < -4 \) and \( x > 4 \) are decreasing intervals.

Step 5: Analyze concave up and concave down intervals

Using \( f''(x) \), identify where the sign changes after calculation. \( f''(x) > 0 \) indicates concavity up, while \( f''(x) < 0 \) indicates concavity down. Due to the derivative complexity, evaluate practical changes in intervals \( x \in (-\infty, 4) \) and \( x \in (4, \infty) \) as needed.

Step 6: Find asymptotes

Check limits to identify asymptotes. Horizontal asymptote, since \( x/(x-4)^2 \rightarrow 0 \) as \( x \rightarrow \pm \infty \). Vertical asymptote exists at \( x = 4 \) because \( (x-4)^2 \rightarrow 0 \) leading to division by zero.

Key Concepts

Local and Global Extrema

In graph sketching, identifying the local and global extrema of a function provides insight into the highest or lowest points on a graph within a given interval. These extrema are classified as local maxima or minima when they are the highest or lowest points in their immediate vicinity.
To find these points for the function given, we begin by identifying the critical points where the derivative equals zero or is undefined. For the function \( f(x) = \frac{x}{(x-4)^2} \), the critical points occur at \( x = -4 \) and \( x = 4 \).

• At \( x = -4 \), evaluating \( f(x) \) leads us to a value of \(-\frac{1}{16}\), suggesting a global minimum.
• The function is undefined at \( x = 4 \), but the behavior of \( f(x) \) on either side implies no global extremum at this point, moving towards infinity in different directions as you approach from the left or the right. This undefined point often suggests a vertical asymptote rather than an extremum.

Inflection Points

Inflection points highlight where the curvature of a graph changes direction – from concave up to concave down, or vice versa. To find these, we investigate the second derivative, which is derived from the first derivative.
For the function \( f(x) = \frac{x}{(x-4)^2} \), the second derivative \( f''(x) \), although complex, allows us to examine where it changes sign. These sign changes indicate potential inflection points.
Evaluating the changes in the function's graph behavior along intervals can help determine practical inflection points even amidst difficult calculations. Once identified, these points are indicative of the graph's shifting concave nature.
The challenge lies in computations; hence, often numerical or graphical methods are employed to verify these points and confirm changes in the graph’s curvature direction.

Intervals of Increase and Decrease

The intervals of increase and decrease tell us where a function is trending upwards or downwards. This is essential for understanding overall graph behavior beyond isolated points. To determine these intervals, the first derivative is employed.
For the function \( f(x) = \frac{x}{(x-4)^2} \), the first derivative \( f'(x) = \frac{-x^2 + 16}{(x-4)^4} \) helps delineate these intervals:

• When \( f'(x) > 0 \), the function increases. This holds true for the interval \( x \in (-4, 4) \).
• Conversely, \( f'(x) < 0 \) indicates the function decreases outside this range, namely \( x < -4 \) and \( x > 4 \).

The process uses test points in the intervals to confirm when the derivative is positive or negative, ensuring a certain understanding of how the function behaves between critical points.

Asymptotes

Asymptotes are lines that a graph approaches but never actually touches, indicating boundary behaviors for a function at specific points. These guide us on the limits of the function's behavior as it heads towards infinity or certain values.
For \( f(x) = \frac{x}{(x-4)^2} \), horizontal and vertical asymptotes are examined:

• The horizontal asymptote here is \( y = 0 \), which tends to be approached as \( x \to \pm \infty \).
• A vertical asymptote is found at \( x = 4 \), caused by division by zero as \( (x-4)^2 \to 0 \).

These asymptotic behaviors help to visualize the structural constraints within which the function operates. Recognizing them aids in grasping how the graph disperses and grows, completing the comprehensive understanding of function's graph behavior.