Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 43-47, the graph of \(y=f(x)\) depends on a parameter \(c\). Using a \(C A S\), investigate how the extremum and inflection points depend on the value of \(c\). Identify the values of \(c\) at which the basic shape of the curve changes. $$ f(x)=\frac{1}{x^{2}+4 x+c} $$

Step-by-Step Solution

Answer

Setting \( c = 4 \) changes the shape at \( x = -2 \). Solving \( 3x^2 + 8x + 4 = c \) shows inflection changes with \( c \).

Step 1: Understand the Function

The given function is \( f(x) = \frac{1}{x^2 + 4x + c} \). We need to explore how the extremum (minima and maxima) and inflection points change as the parameter \( c \) changes.

Step 2: Find the First Derivative

The first derivative of \( f(x) \), \( f'(x) \), is used to find extremum points. Using calculus, calculate the derivative:\[ f'(x) = -\frac{2x + 4}{(x^2 + 4x + c)^2} \].

Step 3: Set the First Derivative to Zero

To find the critical points, set \( f'(x) = 0 \):\[ -\frac{2x + 4}{(x^2 + 4x + c)^2} = 0 \].This leads to solving \( 2x + 4 = 0 \), which gives \( x = -2 \).

Step 4: Evaluate the Extremum

Plug \( x = -2 \) into \( f(x) \) to find the extremum:\[ f(-2) = \frac{1}{(-2)^2 + 4(-2) + c} = \frac{1}{4 - 8 + c} = \frac{1}{c - 4} \].For \( c = 4 \), note that \( f(x) \) becomes undefined at \( x = -2 \).

Step 5: Find the Second Derivative

To analyze inflection points, find the second derivative \( f''(x) \), which is:\[ f''(x) = \frac{2(3x^2 + 8x + 4 - c)}{(x^2 + 4x + c)^3} \].

Step 6: Set Second Derivative to Zero

For inflection points, set \( f''(x) = 0 \):\[ 2(3x^2 + 8x + 4 - c) = 0 \].Solving \( 3x^2 + 8x + 4 - c = 0 \), gives the roots as potential inflection points.

Step 7: Explore How c Changes Shape

The critical points and zeros of the offset equation \( 3x^2 + 8x + 4 - c = 0 \) will indicate shifts in function shape. The discriminant of \( 3x^2 + 8x + 4 = c \), \( 64 - 48 \times 16c \), determines when roots switch by determining where it equals zero. This finds transition \( c \) values.

Key Concepts

Extremum Points

In calculus, extremum points refer to locations on a graph where the function reaches either a local maximum or minimum. These are the critical points of the function, where the slope is zero, meaning the first derivative equals zero.

For the given function, \( f(x) = \frac{1}{x^2 + 4x + c} \), finding extremum points involves:

• Taking the first derivative \( f'(x) \).
• Setting \( f'(x) = 0 \) to solve for \( x \).

In this exercise, solving \( -\frac{2x + 4}{(x^2 + 4x + c)^2} = 0 \) leads to \( x = -2 \).

At \( x = -2 \), plugging into the original function yields the extremum value, \( f(-2) = \frac{1}{c - 4} \). As \( c \) changes, the existence and nature of the extremum can alter significantly.

• If \( c = 4 \), the function is undefined at \( x = -2 \), indicating a potential vertical asymptote.
• For \( c > 4 \) or \( c < 4 \), \( x = -2 \) is a valid extremum point.

Understanding these points helps us predict when the function changes its increasing or decreasing behavior, guided by the parameter \( c \).

Inflection Points

Inflection points are key to determining where a function changes its concavity, transitioning between concave up and concave down. This occurs where the second derivative changes sign.

The second derivative for our function is \[ f''(x) = \frac{2(3x^2 + 8x + 4 - c)}{(x^2 + 4x + c)^3} \].To find the inflection points, we:

• Set \( f''(x) = 0 \) and solve \( 3x^2 + 8x + 4 - c = 0 \).

The solutions to this equation are the x-values of potential inflection points. These roots depend on the discriminant of the quadratic equation.

• A positive discriminant means two real roots, indicating two possible inflection points.
• Zero discriminant means one real root, a transition in concavity at that singular point.
• A negative discriminant implies no real inflection points within the domain.

By examining where the second derivative equals zero, you can understand how \( c \) influences the concavity changes in \( f(x) \). These points are crucial for understanding the curve's shape and behavior.

Graphical Behavior

Understanding the graphical behavior involves integrating knowledge of both extremum and inflection points to grasp how the graph of \( f(x) \) behaves as a whole. It's influenced heavily by parameter \( c \), which adjusts the function's denominator, driving changes in shape and other behaviors.

As the value of \( c \) varies, observe:

• The location and nature of extremum points can shift, affecting the graph's peaks and valleys.
• Inflection points will adjust, informing you about changes in curvature and smooth transitions between concavity.
• Vertical asymptotes may appear, particularly when \( c = 4 \), leading to undefined points and a drastic change in the graph.

Tracking these changes is essential for understanding different configurations or 'shapes' of \( f(x) \). The discriminant \( 64 - 48 \times 16c \) plays a crucial role in identifying transition points where the overall behavior of the graph changes markedly.

By examining \( c \), students can predict shifts in the graph's behavior and anticipate how the curve might look for different values, enhancing comprehension of its dynamical nature.