Problem 48 - Thomas Calculus

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

The local and absolute minimum is at \(x = -1\) with \(k(-1) = 0\).

Step 1: Find the derivative of the function

To find the local extreme values, we first need to find the derivative of the function \(k(x) = x^3 + 3x^2 + 3x + 1\). The derivative \(k'(x)\) will help us identify critical points.Differentiating term-by-term, we have:\[ k'(x) = 3x^2 + 6x + 3\]

Step 2: Find critical points

Critical points occur where the derivative is zero or undefined. Since \(k'(x)\) is a polynomial, it is defined everywhere, so we solve \(k'(x) = 0\):\[3x^2 + 6x + 3 = 0\]Divide the entire equation by 3:\[x^2 + 2x + 1 = 0\]This is a perfect square trinomial. Factor it:\[(x + 1)^2 = 0\]So, \(x = -1\) is the only critical point.

Step 3: Evaluate the function at critical points and endpoints

The critical point \(x = -1\) is within the given domain \((-\infty, 0]\). Now, evaluate \(k(x)\) at \(x = -1\) and at the right endpoint \(x = 0\).For \(x = -1\): \[k(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0\]For \(x = 0\):\[k(0) = (0)^3 + 3(0)^2 + 3(0) + 1 = 1\]

Step 4: Determine local extrema

The value at \(x = -1\) is \(k(-1) = 0\), and the value at the endpoint \(x = 0\) is \(k(0) = 1\). Since \(k(-1)\) is less than \(k(0)\), \(x = -1\) is a local minimum within the interval.

Step 5: Determine absolute extrema

Since the function is defined on the interval \((-\infty, 0]\), we only look within this range. The function decreases leftwards as observed from its derivative behavior, suggesting no lower value exists before \(x=-1\). Since \(k(-1) = 0\) and is less than \(k(0) = 1\), \(x = -1\) is also an absolute minimum.

Step 6: Graphing the function

Using a graphing calculator or graphing tool, plot \(k(x) = x^3 + 3x^2 + 3x + 1\) for the interval \((-\infty, 0]\). The graph should confirm that the function decreases until \(x = -1\) and rises thereafter, showing \(x = -1\) as the lowest point in the interval.

Key Concepts

Derivatives

Derivatives play a crucial role in calculus, especially in finding critical points and understanding how a function behaves. The derivative of a function provides the rate at which the function's value is changing at any given point. In essence, it's like a snapshot of the function's slope at a specific location.

• The derivative, noted as \( k'(x) \), helps us determine where the function increases or decreases.
• To find the derivative of a polynomial function, differentiate each term separately.

For the function given, \( k(x) = x^3 + 3x^2 + 3x + 1 \), the derivative is calculated as follows:

• Derivative of \( x^3 \) is \( 3x^2 \)
• Derivative of \( 3x^2 \) is \( 6x \)
• Derivative of \( 3x \) is \( 3 \)
• Derivative of the constant 1 is 0

Put together, \( k'(x) = 3x^2 + 6x + 3 \), which will guide us to find critical points.

Critical Points

Critical points are where the action happens in calculus. They are important because they show where a function's derivative is zero or doesn't exist. These points help identify potential peaks, troughs, or flat areas.

• To find them, solve \( k'(x) = 0 \).

For our derivative \( 3x^2 + 6x + 3 \), set it to zero and solve: \[ 3x^2 + 6x + 3 = 0 \]
Simplifying by dividing all terms by 3 gives:\[ x^2 + 2x + 1 = 0 \]
This is a perfect square trinomial that can be factored to:\[ (x + 1)^2 = 0 \]
So, \( x = -1 \) is the critical point here. It's where the derivative changes, telling us more about the shape of the original function.

Local Extrema

Local extrema refer to the local minimums and maximums in an interval. These are points where a function reaches a local peak or dip compared to nearby values.

• Local extrema are often found at critical points but not always.

In our exercise, we evaluated the function \( k(x) \) at the critical point and an endpoint. For \( x = -1 \):\[ k(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 = 0 \]
And for the endpoint \( x = 0 \):\[ k(0) = 1 \]
Since \( k(-1) \) is less than \( k(0) \), this means \( x = -1 \) is a local minimum within the specified domain \((-fty, 0]\). This understanding of where the local minimum occurs helps locate dips in graphing.

Absolute Extrema

Absolute extrema are the global highs and lows of a function over a given domain. These are the absolute maximum and minimum values that the function achieves.

• To determine absolute extrema, examine the function at all critical points and endpoints.

In this case, the values we have are \( k(-1) = 0 \) and \( k(0) = 1 \). Since we're observing the interval \((-fty, 0]\), and the function trends downward from infinity past \( x = -1 \) to then rise toward \( x = 0 \),

• \( x = -1 \) provides the absolute minimum because \( k(-1) \) is the lowest value in this interval.

Thus, no value is smaller than \( k(-1) \) in the domain considered, confirming an absolute minimum at this point.