Problem 52 - University Calculus: Early Transcendentals

Question

a. Identify the function's local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute? c. Support your findings with a graphing calculator or computer grapher. $$k(x)=x^{3}+3 x^{2}+3 x+1, \quad-\infty

Step-by-Step Solution

Answer

Local min at \( x = -1 \), local max at \( x = 0 \). Absolute min is \( 0 \), absolute max is \( 1 \).

Step 1: Determine Critical Points

To find local extreme values, we first need to compute the derivative of the function. The function given is \( k(x) = x^3 + 3x^2 + 3x + 1 \). The derivative is \( k'(x) = 3x^2 + 6x + 3 \). Set \( k'(x) = 0 \) to find the critical points: \( 3x^2 + 6x + 3 = 0 \).Solve using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 6 \), \( c = 3 \). This gives:\[ x = \frac{-6 \pm \sqrt{36 - 36}}{6} = \frac{-6}{6} = -1 \].The critical point is \( x = -1 \).

Step 2: Evaluate Function at Critical and Endpoint

To identify local extreme values, evaluate the function at the critical point and the endpoint of the domain. Calculate \( k(-1) \):\[ k(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0 \].Next, calculate \( k(0) \) at the endpoint:\[ k(0) = (0)^3 + 3(0)^2 + 3(0) + 1 = 1 \].

Step 3: Determine Type of Extreme Values

Compare the values obtained at the critical point and endpoint. At \( x = -1 \), the value is \( k(-1) = 0 \). At the endpoint \( x = 0 \), the value is \( k(0) = 1 \).Since \( k(x) = x^3 + 3x^2 + 3x + 1 \) is defined only for \( x \leq 0 \), and decreases until \( x=-1 \), \( k(-1)=0 \) is a local minimum. Since \( k(x) \) increases to the endpoint, \( k(0)=1 \) is a maximum value in the given domain.

Step 4: Identify Absolute Extreme Values

Based on the domain and the behavior of the function, \( k(-1) = 0 \) is the absolute minimum and \( k(0) = 1 \) is the absolute maximum in the defined domain. This is due to no other values in the domain giving outputs lower than \( 0 \) or higher than \( 1 \).

Step 5: Verify with Graph

Use a graphing calculator or computer grapher to visualize \( k(x) = x^3 + 3x^2 + 3x + 1 \) over the domain \(-\infty < x \leq 0\). The graph should show the function decreasing to a minimum at \( x = -1 \), then increasing to a maximum at \( x = 0 \), confirming our previous analysis.

Key Concepts

Critical Points

Critical points are where the derivative of a function is zero or undefined, and they play a crucial role in identifying extreme values in calculus. For the function given, \( k(x) = x^3 + 3x^2 + 3x + 1 \), finding the critical points begins with computing the derivative. The differentiation results in \( k'(x) = 3x^2 + 6x + 3 \). Setting this equal to zero helps us find potential critical points:

• Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 6 \), and \( c = 3 \).
• Solve the equation \( 3x^2 + 6x + 3 = 0 \) to find \( x = -1 \).

Thus, \( x = -1 \) is identified as a critical point for the function. Understanding critical points helps in determining where to examine function behavior closely, checking for extremities like local maxima or minima.

Local Extreme Values

Local extreme values are the points where a function reaches local peaks (maxima) or valleys (minima). Once the critical points are found, like \( x = -1 \) for our function \( k(x) = x^3 + 3x^2 + 3x + 1 \), we need to evaluate the function at these points and possibly at endpoints of the given domain to understand its behavior. Let's do that:

• Evaluate at the critical point, \( k(-1) = 0 \).
• Evaluate at the endpoint, \( k(0) = 1 \).

Since \( k(-1) = 0 \) is a lower point compared to the other values in its immediate neighborhood within the domain, it's a local minimum. Conversely, \( k(0)=1 \) is higher than values nearby in the domain, crowning it a local maximum. Through evaluating these key values, local extrema highlight how critical points and endpoints affect the function's shape over the specified domain.

Absolute Extreme Values

Absolute extreme values are the highest and lowest points in the entire domain of a function. To find these, compare the function's output at critical points and endpoints. Our goal is to determine which of these local extreme values can be absolute within the given domain.

• Check the outputs: \( k(-1) = 0 \) and \( k(0) = 1 \).
• Within the domain \( -\infty < x \leq 0 \), see if any values are less than 0 or greater than 1.

No values are found beyond these boundaries within the domain. Hence, \( k(-1) = 0 \) is the absolute minimum and \( k(0) = 1 \) is the absolute maximum. It is confirmed on visualization—where a graph illustrates the function touching these extreme values—ensuring that no hidden extraneous global extreme values are bypassed. Understanding absolute extremes is imperative for grasping a function's overall highest and lowest points over its range, which is frequently a vital goal in calculus.