Problem 18 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(17-40\) , (a) find the critical numbers of \(f\) (if any), (b) find the open interval(s) on which the function is increasing or decreasing, (c) apply the First Derivative Test to identify all relative extrema, and (d) use a graphing utility to confirm your results. $$ f(x)=x^{2}+6 x+10 $$

Step-by-Step Solution

Answer

The function \(f(x) = x^{2} + 6x + 10\) has a critical number at \(x = -3\), is decreasing on the interval \(-\infty < x < -3\) and is increasing on the interval \(-3 < x < +\infty\). The function has a relative minimum at \(x = -3\).

Step 1: Differentiate the Function

The derivative of \(f(x) = x^{2} + 6x + 10\) is calculated using the power rule for differentiation which states that the derivative of \(x^n\) is \(nx^{n-1}\). Hence the derivative \(f'(x)\) is found as \(2x + 6\).

Step 2: Find Critical Numbers

Critical numbers are the points where the derivative equals zero or the derivative does not exist. For this function, there are no points where the derivative does not exist so setting the derivative equal to zero will give the critical points. Solving \(2x + 6 = 0\) yields \(x = -3\) as the critical point.

Step 3: Find Intervals of Increasing and Decreasing

The intervals of increasing and decreasing can be found by examining the sign of the derivative on the intervals split by the critical numbers. The number line is divided into two intervals by the critical number \(x = -3\): \(-\infty < x < -3\) and \(-3 < x < +\infty\). By picking a test point in each interval, \(x = -4\) in the first interval and \(x = 0\) in the second, and plugging these points into the derivative, one can find the sign of the derivative in each interval. It is found that \(f'(-4) = -2 < 0\) and thus the function is decreasing on the interval \(-\infty < x < -3\), and \(f'(0) = 6 > 0\) so the function is increasing on the interval \(-3 < x < +\infty\).

Step 4: Apply the First Derivative Test

According to the first derivative test, if the derivative of a function changes sign from negative to positive at a critical number, then that critical number is a relative minimum of the function. Here, the derivative changes sign from negative to positive at \(x = -3\), so the function has a relative minimum at \(x = -3\).

Step 5: Verification Using a Graphing Utility

The results can be verified by graphing the function using a graphing calculator or software. The graph should show a clear minimum at \(x = -3\), and the function should be decreasing to the left and increasing to the right of that point.

Key Concepts

Critical Numbers

Critical numbers in a function are crucial for understanding the function's behavior and finding where potential maxima and minima occur. These are the values of \(x\) where the derivative \(f'(x)\) is zero or undefined. To find them, follow these steps:

• First, take the derivative of the function. For the function \(f(x) = x^2 + 6x + 10\), the derivative using the power rule is \(f'(x) = 2x + 6\).
• Set the derivative equal to zero to solve for \(x\). For this example, the equation is \(2x + 6 = 0\). Solving this gives \(x = -3\).
• Check if the derivative is undefined at any point, which isn’t the case here since \(f'(x) = 2x + 6\) is a polynomial and hence, defined everywhere.

Identifying these critical numbers helps in analyzing a function's graph, highlighting where changes in direction occur.

Increasing and Decreasing Intervals

Understanding where a function increases or decreases provides insight into its behavior. These intervals are determined by the sign of the derivative:

• Once you've found the critical number \(x = -3\), divide the number line into intervals based on this critical point: \(-\infty < x < -3\) and \(-3 < x < +\infty\).
• Select test points from each interval and substitute these into the derivative \(f'(x) = 2x + 6\). A test point \(x = -4\) yields \(f'(-4) = -2\), indicating a decreasing interval \((-\infty, -3)\) as the derivative is negative.
• Similarly, a test point \(x = 0\) gives \(f'(0) = 6\), indicating an increasing interval \((-3, \infty)\) as the derivative is positive.

This simple analysis helps to sketch or understand the slope and curve behavior of the function over its domain.

Relative Extrema

Relative extrema are points where the function reaches local maxima or minima. The First Derivative Test is useful for identifying these extrema:

• If a function's derivative changes from negative to positive at a critical point, this indicates a relative minimum.
• For the function \(f(x) = x^2 + 6x + 10\), at \(x = -3\), the derivative changes sign from negative to positive. Thus, \(x = -3\) is a relative minimum.
• To find the actual value of the minimum point, plug the critical number \(x = -3\) back into the original function. This yields \(f(-3) = (-3)^2 + 6(-3) + 10 = 1\).

Therefore, the relative minimum of the function is at the point \((-3, 1)\), understood and verified through the First Derivative Test.

Power Rule for Differentiation

The Power Rule for Differentiation is a foundational tool for solving calculus problems, making it simple to find derivatives of functions of the form \(x^n\):

• This rule states: if you have \(x^n\), the derivative is \(nx^{n-1}\).
• In our example function \(f(x) = x^2 + 6x + 10\), apply the power rule: the derivative of \(x^2\) is \(2x^{2-1} = 2x\), and the derivative of \(6x\) is \(6x^{1-1} = 6\).
• Combined, the derivative is \(f'(x) = 2x + 6\), a necessary step to find critical numbers and analyze the function's behavior.

Utilizing the power rule simplifies calculus computations, reducing complex polynomial derivatives to a straightforward task.