Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 1 through 12: (a) Find all critical points. (b) Find \(f^{\prime \prime} .\) Use the second derivative, wherever possible, to determine which critical points are local maxima and which are local minima. If the second derivative test fails or is inapplicable, explain why and use an alternative method for classifying the critical point. $$ f(x)=x^{6}+x^{4} $$

Step-by-Step Solution

Answer

The critical point is at \(x=0\). The second derivative test fails at \(x=0\), as \(f''(0)=0\), which is inconclusive. However, by observing the non-negativity of the function for all real x, we can conclude that \(x=0\) is a global minimum, and thus, a local minimum as well.

Step 1: Calculate the first derivative

We differentiate \(f(x) = x^{6} + x^{4}\) with respect to x. The derivative, denoted as \(f'(x)\), is given by \(f'(x) = 6x^{5} + 4x^{3}\).

Step 2: Find the critical points

We set the first derivative equal to zero to find the critical points: \(0 = 6x^{5} + 4x^{3}\). Factoring out \(2x^{3}\) gives \(2x^{3}(3x^{2}+2) = 0\). Setting each factor equal to zero yields the critical points \(x = 0, \sqrt{-\frac{2}{3}}\), and \(-\sqrt{-\frac{2}{3}}\). However, since the square root of a negative number is not a real number, the only real critical point is \(x=0\).

Step 3: Calculate the second derivative

The second derivative, denoted as \(f''(x)\), can be obtained by differentiating \(f'(x) = 6x^{5} + 4x^{3}\) with respect to x: \(f''(x) = 30x^{4} + 12x^{2}\).

Step 4: Classify the critical point using the second derivative test

Substituting \(x=0\) into \(f''(x)\) gives \(f''(0)= 0\). The second derivative test is inconclusive when \(f''(x)\) equals zero or is undefined, so the second derivative test fails to determine the nature of the critical point at \(x=0\). In such scenarios, an alternative method should be used. Here one can observe that the given function \(f(x) = x^{6} + x^{4}\) is always non-negative for all real x, since both \(x^{6}\) and \(x^{4}\) are always non-negative. Consequently, \(x=0\) must be a global minimum.

Key Concepts

Second Derivative Test

The second derivative test is a handy tool to determine if a critical point is a local maximum or a local minimum. Once you've found a critical point by setting the first derivative to zero, you calculate the second derivative of the function and plug the critical point into it.

• If the second derivative is positive at that point, the function has a local minimum there because the graph is concave up.
• If the second derivative is negative, the function has a local maximum because the graph is concave down.
• If the second derivative is zero, the test is inconclusive. You have to use another method to determine the nature of the critical point in this case.

In our exercise, when we substituted the critical point into the second derivative, we got zero, meaning the test failed. When this happens, always consider other aspects of the function, such as its domain or continuity, to analyze further attributes like global minima or maxima.

First Derivative

The first derivative is the tool we use to find critical points in a function. When you take the derivative of a function, what you're actually finding is the slope of the function at any given point.

• When the first derivative equals zero, the slope is zero, indicating a potential critical point. These are places where the function's rate of change is zero, which means the function could change from increasing to decreasing, or vice versa.
• Critical points can be where the function achieves a local maximum, a local minimum, or simply a point of inflection.

For the provided function, the derivative was calculated by standard differentiation rules. Once we had the derivative, setting it to zero allowed us to find the critical points. However, when dealing with polynomials, not every solution to the derivative will be real numbers, as seen in our exercise.

Local Maxima

Local maxima are points on the graph of a function where the value is higher than any nearby points. They are the "peaks" that aren't necessarily the highest point overall but are higher than the immediate surroundings.
To identify a local maximum:

• Use the first derivative to find critical points.
• Apply the second derivative test. If the second derivative at a critical point is negative, then there's a local maximum.
• If the second derivative test is inconclusive, check intervals around the critical point or use other methods.

In the given exercise, the function did not have a local maximum at the real critical point. This is because the second derivative test returned zero, and with further observation, the function forms a global minimum at that point.

Local Minima

Local minima are quite the opposite of local maxima. Here, the function reaches a lower value than any other nearby points, forming a 'valley' in the graph. They aren't necessarily the lowest point worldwide but lower than their immediate surroundings.
To identify a local minimum:

• Begin by calculating the critical points using the first derivative.
• Use the second derivative test, and if it's positive at a critical point, that point is a local minimum.
• If the second derivative test fails, such as when it is zero, inspect the function's graph or use other methods to ascertain the nature of the point.

Within the discussed exercise, we applied this to the critical point and inferred that the shape of the function was bottomed out there. The graph of the function further elucidated this point, showing the function remaining non-negative across real numbers, thus confirming the critical point was a global minimum.