Problem 39 - Calculus Early Transcendentals

Question

a. Locate the critical points of \(f\) b. Use the First Derivative Test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist). $$f(x)=x^{2}+3 \text { on }[-3,2]$$

Step-by-Step Solution

Answer

Answer: The absolute maximum value is 12 at \(x=-3\), and the absolute minimum value is 3 at \(x=0\).

Step 1: Find the first derivative of the function \(f(x)\)

To find the first derivative, apply the power rule to each term: $$f'(x) = (x^2)' + (3)' = 2x + 0 = 2x$$

Step 2: Determine the critical points

Critical points occur where the first derivative is equal to zero or is undefined. Since the function \(f'(x)=2x\) is defined for all x-values, we only need to find the x-values when it is equal to zero: $$2x = 0 \Rightarrow x=0$$ So we have only one critical point, \(x=0\).

Step 3: Use the First Derivative Test

Using the First Derivative Test, we'll evaluate the sign of \(f'(x)\) to the left and right of the critical point \(x=0\): - For \(x<0\): \(f'(x) = 2x < 0\), so \(f(x)\) is decreasing. - For \(x>0\): \(f'(x) = 2x > 0\), so \(f(x)\) is increasing. From the above information, we can conclude that the critical point \(x=0\) gives us a local minimum value.

Step 4: Identify the absolute maximum and minimum values on the interval \([-3, 2]\)

To find the absolute maximum and minimum values, we must evaluate the given function \(f(x) = x^2 + 3\) at the critical point and the endpoints of the interval. 1. For \(x=-3\): \(f(-3) = (-3)^2 + 3 = 9 + 3 = 12\) 2. For \(x=0\) (critical point): \(f(0) = 0^2 + 3 = 3\) 3. For \(x=2\): \(f(2) = 2^2 + 3 = 4 + 3 = 7\) Looking at the function values, we can identify the following: - The absolute maximum value is 12 at \(x=-3\). - The absolute minimum value is 3 at \(x=0\).

Key Concepts

First Derivative Test

Grasping the First Derivative Test is key to understanding where a function reaches its high and low points. Begin with the derivative of a function, and then locate its critical points by solving for when the derivative equals zero. To apply the test, analyze the sign of the derivative—positive or negative—on either side of each critical point.

For example, if the derivative shifts from negative to positive as you pass through a critical point, the function moved from decreasing to increasing, pinpointing a local minimum. Conversely, the derivative turning from positive to negative suggests you've found a local maximum at the critical point. This method confidently classifies local extremas, pivotal in deciphering the behavior of functions.

Absolute Maximum and Minimum Values

Locating the absolute maximum and minimum values of a function goes one step further than finding local extremes. The task is to determine the highest and the lowest points across the entire interval. The contenders for these absolute values include the local extremes and the endpoint values of the interval.

To ensure you've vetted all possible candidates, evaluate the function at the critical points as well as at the start and end points of the interval. The largest and smallest results are the absolute maximum and minimum, respectively. These values give us an encompassing view of the function's reach over the specified domain.

Power Rule Differentiation

The power rule is a fundamental technique for differentiating functions with powers of x. It states that if you have a function of the form \(f(x) = x^n\), its derivative \(f'(x)\) is \(nx^{n-1}\). This rule simplifies the process of differentiation by quickly generating the derivative without complex calculations.

Using this principle, we can swiftly find the derivative of \(x^2\) as \(2x^{2-1} = 2x\). The power rule is a staple in calculus, paving the way for identifying critical points and aiding in the analysis of a function's rising and falling behavior.

Interval Evaluation

Interval evaluation involves assessing a function's behavior within specific boundaries, or intervals, particularly when seeking absolute extremas. Beyond finding the critical points, check the function's value at the interval's endpoints because they can be pivotal in pinpointing the absolute extremes.

This is accomplished by simply substituting the endpoints into the function and comparing the resulting values with those obtained at the critical points. In essence, interval evaluation ensures no stone is left unturned in the quest for the most comprehensive assessment of a function's behavior within a specified range.