Problem 118 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Consider the functions \(f(x)=x^{3}, g(x)=x^{4}\), and \(h(x)=-x^{4}\) a. Show that \(x=0\) is a critical number of each of the functions \(f, g\), and \(h .\) b. Show that the second derivative of each of the functions \(f, g\), and \(h\) equals zero at \(x=0\). c. Show that \(f\) has neither a relative maximum nor a relative minimum at \(x=0\), that \(g\) has a relative minimum at \(x=0\), and that \(h\) has a relative maximum at \(x=0\).

Step-by-Step Solution

Answer

In summary, for the functions \(f(x) = x^3\), \(g(x) = x^4\), and \(h(x) = -x^4\), the critical number \(x = 0\) is found for each of them, and their second derivatives are all equal to zero at \(x = 0\). However, the second derivative test is inconclusive for these cases, so we use the graphs to determine that \(f(x) = x^3\) has neither a relative maximum nor minimum at \(x = 0\), \(g(x) = x^4\) has a relative minimum at \(x = 0\), and \(h(x) = -x^4\) has a relative maximum at \(x = 0\).

Step 1: Finding critical numbers for each function

To find the critical numbers for each function, we first need to find the first derivative of each function and then find where the derivative is equal to zero. 1. \(f(x) = x^3\) Derivative: \(f'(x) = 3x^2\). Set the derivative equal to zero: \(3x^2 = 0\). Solve for \(x\): \(x = 0\). 2. \(g(x) = x^4\) Derivative: \(g'(x) = 4x^3\). Set the derivative equal to zero: \(4x^3 = 0\). Solve for \(x\): \(x = 0\). 3. \(h(x) = -x^4\) Derivative: \(h'(x) = -4x^3\). Set the derivative equal to zero: \(-4x^3 = 0\). Solve for \(x\): \(x = 0\). So, \(x = 0\) is a critical number for each of the functions \(f, g\), and \(h\).

Step 2: Finding the second derivative for each function

Now we need to find the second derivative of each function and evaluate it at \(x = 0\). 1. \(f'(x) = 3x^2\) Second derivative: \(f''(x) = 6x\). Evaluate at \(x = 0\): \(f''(0) = 6(0) = 0\). 2. \(g'(x) = 4x^3\) Second derivative: \(g''(x) = 12x^2\). Evaluate at \(x = 0\): \(g''(0) = 12(0)^2 = 0\). 3. \(h'(x) = -4x^3\) Second derivative: \(h''(x) = -12x^2\). Evaluate at \(x = 0\): \(h''(0) = -12(0)^2 = 0\). The second derivative of each of the functions \(f, g\), and \(h\) equals zero at \(x = 0\).

Step 3: Determining relative maxima and minima

We will use the second derivative test to determine whether there are relative maxima, minima, or neither. 1. For \(f(x) = x^3\), \(f''(x) = 6x\) and \(f''(0) = 0\). Since the second derivative is zero, the second derivative test is inconclusive. By looking at the graph of \(f(x) = x^3\), we can tell that there is no minimum or maximum at \(x = 0\). 2. For \(g(x) = x^4\), \(g''(x) = 12x^2\) and \(g''(0) = 0\). Although the second derivative is inconclusive as it's zero, since \(g'(x) = 4x^3\) and the function is an even function, the region to the left and right of \(x = 0\) is symmetric. We can infer there is a relative minimum at \(x = 0\). 3. For \(h(x) = -x^4\), \(h''(x) = -12x^2\) and \(h''(0) = 0\). Similar to the case for \(g(x)\), although the second derivative is inconclusive as it's zero, since \(h'(x) = -4x^3\) and it's also an even function, the region to the left and right of \(x = 0\) is symmetric. We can infer there is a relative maximum at \(x = 0\). In conclusion, \(f(x) = x^3\) has neither a relative maximum nor minimum at \(x = 0\), \(g(x) = x^4\) has a relative minimum at \(x = 0\), and \(h(x) = -x^4\) has a relative maximum at \(x = 0\).

Key Concepts

Second Derivative Test

The second derivative test is a useful tool in calculus for determining the nature of critical points. It helps to ascertain whether a given critical point is a point of relative maximum, relative minimum, or if the test is inconclusive. Here's how it works:

• First, you must find the first derivative of a function and identify the critical points by setting the first derivative equal to zero.
• Next, take the second derivative of the original function.
• Evaluate the second derivative at each critical point:
• If the second derivative is positive (\(f''(x) > 0\)) at a critical point, the function has a relative minimum at that point.
• If the second derivative is negative (\(f''(x) < 0\)) at a critical point, the function has a relative maximum at that point.
• If the second derivative is zero (\(f''(x) = 0\)), the test is inconclusive, and other methods or information (such as analyzing the graph) should be used to determine the nature of the critical point.

For example, in the given exercise with functions \(f(x)=x^3\), \(g(x)=x^4\), and \(h(x)=-x^4\), the second derivative test at \(x=0\) was inconclusive as \(f''(0)=0\), \(g''(0)=0\), and \(h''(0)=0\). Thus, further analysis was necessary.

Relative Maximum

A relative maximum is a point on the graph of a function where the function value is higher than at any nearby points. This implies that, within a small interval around this point, the function does not reach a higher value.

• To identify a relative maximum using derivatives, you typically use the first and second derivatives as explained in the second derivative test.
• If the first derivative changes from positive to negative at a point, or if the second derivative is negative at that critical point, a relative maximum occurs.
• For periodic and smooth functions, like polynomials, these points can often be found and confirmed using simple algebraic methods or by examining the function graphically.

In the exercise, \(h(x)=-x^4\) was examined, and although the second derivative was zero at \(x=0\), symmetry and graph analysis showed \(x=0\) as a relative maximum.

Relative Minimum

A relative minimum is a point on a function's graph where the function value is lower than at any nearby points. It means within a small interval around this point, the function doesn't go lower.

• Generally, if the first derivative changes from negative to positive or if the second derivative is positive at a critical point, there is a relative minimum.
• These points are key in understanding the behavior of functions, particularly in analyses related to optimization and graph interpretation.
• In the context of polynomial functions, as seen with \(g(x)=x^4\), checking for a relative minimum at \(x=0\) required evaluating behavior around this point.

For the polynomial \(g(x)=x^4\), despite the second derivative test being inconclusive, the homogeneous construction and symmetry of the function showed \(x=0\) to indeed be a point of relative minimum.

Polynomial Functions

Polynomial functions are mathematical expressions involving sums of powers in one or more variables multiplied by coefficients. They are fundamental in calculus due to their smooth, continuous nature, making them easy to differentiate and integrate.

• Polynomials of degree \(n\) can have up to \(n−1\) critical points and \(n−2\) inflection points.
• Differentiation of polynomials is straightforward by applying the power rule which reduces terms power by one and multiplies by the original power as a coefficient.
• Critical points of polynomial functions can be easily found by setting the first derivative to zero.

The functions in the exercise, such as \(f(x)=x^3\), \(g(x)=x^4\), and \(h(x)=-x^4\), are simple polynomials. Given their structure, finding critical points and determining their nature through derivatives is an approachable task, illustrating key calculus concepts.