Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For each function: (a) Find all critical points on the specified interval. (b) Classify each critical point: Is it a local maximum, a local minimum, an absolute maximum, or an absolute minimum? (c) If the function attains an absolute maximum and/or minimum on the specified interval, what is the maximum and/or minimum value? \(f(x)=-2 x^{3}+3 x^{2}+12 x+5\) on \((-\infty, \infty)\)

Step-by-Step Solution

Answer

The function has critical points at \(x = 2\) and \(x = -1\). \(x = 2\) corresponds to a local and absolute maximum with a value of 9, and \(x = -1\) corresponds to a local and absolute minimum with a value of -6.

Step 1: Find the first derivative of the function

To find the critical points, we need to first find the derivative of the function. The derivative of the function \(f(x)=-2x^{3}+3x^{2}+12x+5\) is \(f'(x) = -6x^{2}+6x+12\).

Step 2: Set the derivative equal to zero and solve for x

The critical points occur where the derivative is zero. Setting \(f'(x) = 0\) gives the equation \(-6x^{2}+6x+12 = 0\). Dividing through by -6, we get \(x^{2}-x-2 = 0\). We can then solve this equation by factoring to get \(x-2)(x+1) = 0\), which gives us the solutions \(x = 2\) and \(x = -1\). These are the critical points of the function.

Step 3: Use the second derivative test to classify the critical points

To classify the critical points, we find the second derivative, \(f''(x) = -12x + 6\), and evaluate it at the critical points. For \(x = 2\), \(f''(2) = -12(2) + 6 = -18 < 0\), so \(x = 2\) is a local maximum. For \(x = -1\), \(f''(-1) = -12(-1) + 6 = 18 > 0\), so \(x = -1\) is a local minimum.

Step 4: Evaluate the function at the critical points and endpoints of the interval

The maximum and minimum of the function on the interval \((-\infty, \infty)\) occur at the critical points or at the endpoints. However, because the interval is \((-\infty, \infty)\), there are no endpoints to evaluate. Therefore, we substitute the critical points into the original function: \(f(2) = -2(2)^{3}+3(2)^{2}+12(2)+5 = 9\) and \(f(-1) = -2(-1)^{3}+3(-1)^{2}+12(-1)+5 = -6\). Hence, the function has a local (and thus absolute) maximum value of 9 at \(x = 2\) and a local (and thus absolute) minimum value of -6 at \(x = -1\).

Key Concepts

First Derivative Test

The first derivative test is a powerful tool in calculus to determine where a function has local maximum or minimum points. To apply this test, we start by finding the function's first derivative and setting it equal to zero to solve for the critical points, which are potential locations for these extrema. In our example, the critical points of the function \(f(x)=-2x^{3}+3x^{2}+12x+5\) were found at \(x = 2\) and \(x = -1\) by setting the first derivative \(f'(x) = -6x^{2}+6x+12\) to zero and solving the resulting quadratic equation.

After identifying the critical points, we analyze the sign of the derivative before and after these points on a number line to determine whether the function is increasing or decreasing. If the derivative changes from positive to negative, we have a local maximum, and if it changes from negative to positive, we have a local minimum. This test gives us insight into the behavior of the function and helps classify the critical points accordingly.

Second Derivative Test

The second derivative test is another method used to classify critical points. With this test, after finding the critical points using the first derivative, we take the second derivative of the function and evaluate it at each critical point. The sign of the second derivative tells us about the concavity of the function at those points. If the second derivative is positive, the function is concave up, indicating a local minimum; if it's negative, the function is concave down, suggesting a local maximum.

In our exercise, we calculated the second derivative \(f''(x) = -12x + 6\) to test the critical points \(x = 2\) and \(x = -1\). At \(x = 2\), the second derivative is negative \(f''(2) = -18\), indicating a local maximum. For \(x = -1\), the second derivative is positive \(f''(-1) = 18\), confirming a local minimum. This test is very useful when the first derivative test is inconclusive or difficult to apply.

Local Maximum and Minimum

Local maximum and minimum points are where a function reaches its highest or lowest value, respectively, within a certain interval around these points. Identifying these points involves finding where the first derivative of a function equals zero or does not exist and then classifying these points using either the first or second derivative tests.

It is important to note that local maxima and minima are not necessarily the highest or lowest points over the entire domain of the function, they are just 'peaks' and 'valleys' in the vicinity of the critical points. Using the example of our function \(f(x)=-2x^{3}+3x^{2}+12x+5\), we've identified local extrema at the critical points: a local maximum at \(x = 2\) and a local minimum at \(x = -1\), based on the first and second derivative tests.

Absolute Maximum and Minimum

In contrast to local extrema, absolute maximum and minimum refer to the highest and lowest values that a function takes on over its entire domain. To find these, you would evaluate the function at critical points and at any end points of the domain if it's a closed interval. However, in cases where the function spans from \(-\infty\) to \(\infty\), the absolute extrema may only occur at the critical points, if at all.

We've determined that the function \(f(x)=-2x^{3}+3x^{2}+12x+5\), over the domain \((-fty, fty)\), attains an absolute maximum of 9 at \(x = 2\) and an absolute minimum of -6 at \(x = -1\), as there are no boundaries to the intervals to consider for additional extrema. The absolute extrema are critical for understanding the overall behavior and range of a function.