Problem 8 - 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. $$ f(x)=(x+1) /(x+2)^{2} $$

Step-by-Step Solution

Answer

Local maximum at \(x = 0\); increases \((-\infty, 0)\), decreases \((0,\infty)\); vertical asymptote at \(x = -2\); horizontal asymptote at \(y = 0\); no inflection points.

Step 1: Find the Critical Points

To identify local and global extrema, we start by finding the critical points. First, derive the function using the quotient rule. The original function is \( f(x) = \frac{x+1}{(x+2)^2} \). The derivative, using the quotient rule \( \frac{u}{v}' = \frac{u'v - uv'}{v^2} \), where \( u = x+1 \) and \( v = (x+2)^2 \), is:\[f'(x) = \frac{(1)(x+2)^2 - (x+1)(2)(x+2)}{(x+2)^4}\]Simplifying this we get:\[f'(x) = \frac{x+2 - 2(x+1)}{(x+2)^3} = \frac{-x}{(x+2)^3}\]Setting \( f'(x) = 0 \), gives \( x = 0 \) as a critical point.

Step 2: Determine Extrema using Second Derivative Test

To determine if the critical point is a local extremum, use the second derivative test. The second derivative is found by differentiating \( f'(x) \):\[f''(x) = \frac{d}{dx}\left(\frac{-x}{(x+2)^3}\right)\]Applying the quotient rule again yields:\[f''(x) = \frac{-(x+2)^3 - (-x)(3)(x+2)^2}{(x+2)^6} = \frac{-x(x+8)-8}{(x+2)^6}\]Evaluating at the critical point \( x = 0 \):\[f''(0) = \frac{-0(0+8)-8}{(0+2)^6} = \frac{-8}{64} = -\frac{1}{8}\]Since \( f''(0) < 0 \), \( x = 0 \) is a local maximum.

Step 3: Find Intervals of Increase and Decrease

Based on the first derivative \( f'(x) = \frac{-x}{(x+2)^3} \):- \( f'(x) > 0 \) when \( -x > 0 \), i.e., \( x < 0 \). The function is increasing on \((-\infty, 0)\).- \( f'(x) < 0 \) when \( -x < 0 \), i.e., \( x > 0 \). The function is decreasing on \((0, \infty)\).Hence, there is a local maximum at \( x = 0 \).

Step 4: Analyze Concavity and Inflection Points

Determine concavity by analyzing the second derivative:- \( f''(x) = \frac{-x(x+8)-8}{(x+2)^6} \)- \( f''(x) > 0 \) for concave up; \( f''(x) < 0 \) for concave down.To find inflection points, solve \( f''(x)=0 \):- Simplified, \( -x^2 - 8x - 8 = 0 \). Solve using the quadratic formula:\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(-1)(-8)}}{2(-1)} = \frac{8 \pm \sqrt{64 - 32}}{-2}\]\[ x = \frac{8 \pm \sqrt{32}}{-2} = \frac{8 \pm 4\sqrt{2}}{-2}\]This yields no real changes in concavity, thus no inflection points. Examine intervals to confirm concavity behavior.

Step 5: Identify Asymptotes

The vertical asymptote occurs at \( x+2 = 0 \) implying \( x = -2 \) because the denominator becomes zero there. For horizontal asymptotes, since the degree of the polynomial in the denominator is higher, \( f(x) = \frac{x+1}{(x+2)^2} \to 0 \) as \( x \to \pm \infty \). So, the horizontal asymptote is \( y = 0 \).

Key Concepts

Critical Points

Critical points are where the derivative of a function equals zero or is undefined. These points are crucial because they can indicate potential local maximums or minimums, where the graph "turns around." To find the critical points for the function given, start by finding the derivative using the quotient rule on the function \( f(x) = \frac{x+1}{(x+2)^2} \). The derivative is \[ f'(x) = \frac{-x}{(x+2)^3} \].Set the derivative equal to zero: \( f'(x) = 0 \) implies \( x = 0 \). Thus, \( x = 0 \) is the only critical point for this function. In the context of rational functions, analyzing these points reveals where the function exhibits significant changes in behavior, like peaks, troughs, or flat areas. Identifying these critical points is a foundational step in sketching the graph, helping to pinpoint where the local extrema might occur.

Second Derivative Test

The second derivative test helps determine the nature of the critical points, specifically whether they are local maxima, minima, or neither. Given the second derivative \[ f''(x) = \frac{-x(x+8)-8}{(x+2)^6} \], we evaluate it at the critical point \( x = 0 \). This calculation yields \[ f''(0) = -\frac{1}{8} \],indicating that at \( x = 0 \), the function has a local maximum, since the second derivative is negative.The second derivative essentially measures the concavity or the curvature of the graph at a certain point. Thus, a negative value means the curve is concave down, reinforcing the identification of a local maximum. This step confirms the critical point’s behavior and solidifies our understanding of the graph's shape around this area.

Asymptotes

Asymptotes are lines that a graph approaches but never reaches. Understanding asymptotes is key in sketching rational functions, like \( f(x) = \frac{x+1}{(x+2)^2} \).**Vertical Asymptotes**

• These occur where the function is undefined—specifically, where the denominator equals zero.
• For this function, the vertical asymptote is at \( x = -2 \), since the denominator becomes zero when \( x+2 = 0 \).

**Horizontal Asymptotes**

• These exist when \( x \to \pm \infty \).
• Here, because the highest power in the denominator exceeds that in the numerator, the horizontal asymptote is \( y = 0 \).

Both types of asymptotes guide the overarching behavior of the function, dictating how it extends to infinity or behaves near certain exclusions, framing the overall graphical sketch.

Concavity and Inflection Points

Concavity refers to how a function's graph curves. **Concave up** suggests the curve looks like a cup "smiling," and **concave down** suggests a "frown." The concavity is identified using the second derivative \( f''(x) = \frac{-x(x+8)-8}{(x+2)^6} \).To examine where the function is concave up or down, consider when

• \( f''(x) > 0 \) for concave up, and
• \( f''(x) < 0 \) for concave down.

Inflection points are where the function's concavity changes from up to down, or vice versa. Solve \( f''(x) = 0 \) (though no real solutions occur in this case) to look for inflection points. In our analysis, the equation \(-x^2 - 8x - 8 = 0\) has no real roots associated with any concavity change; hence, there are no inflection points in this example. Understanding concavity and inflection is key as they reveal how the graph behaves over intervals, which supports a deeper comprehension of the function's nature throughout its domain.