Problem 13 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find and classify all critical points. (a) \(f(x)=-x^{3}-3 x^{2}+9 x+5\) (b) \(f(x)=x^{3}+3 x^{2}+9 x+8\)

Step-by-Step Solution

Answer

The critical points for the first function \(f(x)=-x^{3}-3 x^{2}+9 x+5\) are \(x = -1\) (neither max nor min) and \(x = 3\) (max), and for the second function \(f(x)=x^{3}+3 x^{2}+9 x+8\), there are no critical points.

Step 1: Find the derivative of the given functions

The derivative of a function can be found using basic calculus rules. Differentiating \(f(x)=-x^{3}-3 x^{2}+9 x+5\) gives \(f'(x)=-3x^{2}-6x+9\). Differentiating \(f(x)=x^{3}+3 x^{2}+9 x+8\) gives \(f'(x)=3x^{2}+6x+9\).

Step 2: Solve for derivative equals to zero

We can find the critical points by solving the derivative equations we found in step 1 equal to zero. Thus for first function we solve \(0 = -3x^{2}-6x+9\) which gives us \(x = -1, 3\). Solving the equation \(0 = 3x^{2}+6x+9\) of the second function gives no solution. Hence, there are no critical points for the second function.

Step 3: Apply the second derivative test

The second derivative test is used to classify the critical points as maximum, minimum or neither. First, we differentiate the first derivative to get second derivatives. For the first function it will be \(f''(x)=-6x-6\). For the second function it will be \(f''(x)=6x+6\). Now apply these values \(x = -1, 3\) from step 2 to the second derivatives respectively. For the first function, the values are \(f''(-1)= 0\) which neither maximum nor minimum and \(f''(3)= -18\) which is maximum at \(x = 3\). For the second function, since there are no critical points found in step 2, there will be no points to test in the second derivative.

Key Concepts

Derivative of a Function

Understanding the derivative of a function is crucial in calculus. It measures how a function's output changes as its input changes—essentially, it's the rate of change or slope of the function at any point. To find the derivative, we use rules of differentiation such as the power rule, product rule, or chain rule, depending on the form of the function.

For example, in the given exercise, the functions are polynomials, and we use the power rule. The power rule states that if you have a term in the form of ax^n, the derivative is anx^(n-1). Applying this rule to f(x)=-x^3-3x^2+9x+5, we calculate the derivative, f'(x)=-3x^2-6x+9. This derivative tells us how f(x) changes with respect to x.

Finding the derivative is the first step in identifying critical points where the function's slope is zero or undefined, which are often associated with peaks, valleys, or inflection points on the graph of the function.

Second Derivative Test

Once we have discovered potential critical points by setting the first derivative equal to zero, we can use the second derivative test to classify these points further. This test involves taking the second derivative of the function, which provides information on the concavity of the function at a given point. If the second derivative at a critical point is positive, the function is concave up, indicating a local minimum. If it's negative, the function is concave down, and we have a local maximum.

For instance, continuing with our exercise, after finding the first derivative, we take another derivative to get f''(x). For f(x)=-x^3-3x^2+9x+5, we get f''(x)=-6x-6. We then evaluate this at the critical points found earlier. If the second derivative doesn't change sign (from positive to negative or vice versa) as we pass through the critical point, or if it is zero, the test is inconclusive. This is what happened at x = -1; since f''(-1) = 0, we cannot conclude whether it is a max or min, but it’s likely an inflection point.

Classifying Critical Points

Classifying critical points is helpful for understanding the overall behavior of a function. After identifying where the first derivative equals zero or does not exist, we classify these points to determine if they correspond to local maxima, local minima, or inflection points. The second derivative test is one method for doing this. However, when the second derivative test is inconclusive, additional methods, such as analyzing the sign of the first derivative around the critical points or using the first derivative test, can be employed.

For example, in our exercise's first case, at x = 3, the second derivative came out negative (f''(3)=-18), which confirms a local maximum at that point. However, no critical points were found for the second function in its domain, meaning there are no local maxima or minima to classify. For a comprehensive understanding of a function's behavior, it is often necessary to create a sign chart or graph the function to visualize where these changes occur. This multi-faceted approach ensures students can determine the nature of critical points with confidence.