Problem 73 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the relative extrema, if any, of each function. Use the second derivative test, if applicable. $$ f(t)=t^{2}-\frac{16}{t} $$

Step-by-Step Solution

Answer

The relative extrema of the function \(f(t) = t^2 - \frac{16}{t}\) is a relative maximum at the point \((2, -4)\).

Step 1: Find the first derivative and critical points

First, we'll take the derivative of \(f(t)\) with respect to \(t\). \[f'(t) = \frac{d}{dt}(t^2 - \frac{16}{t})\] Now we need to apply the power rule and the chain rule: \[f'(t) = 2t + 16t^{-2}\] To find the critical points, we set the first derivative equal to zero, and solve for \(t\): \[0 = 2t + 16t^{-2}\]

Step 2: Find the second derivative of the function

Now we need to find the second derivative to apply the second derivative test. We'll find the derivative of the first derivative: \[f''(t) = \frac{d}{dt}(2t + 16t^{-2})\] Apply the power rule and the chain rule: \[f''(t) = 2 - 32t^{-3}\]

Step 3: Evaluate the second derivative at the critical points

Before we can evaluate the second derivative at the critical points, we need to find the critical points. Let's rewrite the first derivative equation from step 1: \[t^3 = 8\] Now, solve for \(t\): \[t = 2\] Now we can evaluate the second derivative at the critical point \(t = 2\): \[f''(2) = 2 - 32(2)^{-3} = 2 - 4 = -2\]

Step 4: Determine the nature of the extrema

Since the second derivative \(f''(2) = -2\) is negative at the critical point \(t = 2\), this means that the function has a relative maximum at \(t = 2\). Therefore, there is a relative maximum at: \[f(2) = 2^2 - \frac{16}{2} = 4 - 8 = -4\] As a result, the relative extrema of the function \(f(t) = t^2 - \frac{16}{t}\) is a relative maximum at the point \((2, -4)\).

Key Concepts

Second Derivative Test

The second derivative test is a handy tool in calculus used to determine the nature of critical points. Critical points are where the first derivative of a function is zero or undefined. Essentially, they are potential candidates for local maxima, minima, or points of inflection.
To apply the second derivative test, you first need to find the second derivative of the function, denoted as \(f''(x)\). Here are the steps you should follow:

• Identify the critical points by setting the first derivative \(f'(x)\) equal to zero and solving for \(x\).
• Compute the second derivative \(f''(x)\).
• Evaluate the second derivative \(f''(x)\) at each critical point.
• Based on \(f''(x)\), determine the nature of each critical point:
  • If \(f''(x) > 0\), the function has a relative minimum at that point.
  • If \(f''(x) < 0\), the function has a relative maximum at that point.
  • If \(f''(x) = 0\), the test is inconclusive, and further analysis is needed.

In our case, we evaluated \(f''(t)\) at the critical point \(t = 2\) and found it to be \(-2\). Since \(f''(2) < 0\), we determined that the function has a relative maximum at this point.

Critical Points

Critical points in calculus are essential in identifying where a function's graph changes direction. They occur where the first derivative is zero or undefined, and they represent potential locations for local extrema or points of inflection. Here’s how they work:

• The critical point is found by solving \(f'(x) = 0\).
• Once you have calculated the derivative, set it equal to zero or see where it's undefined.
• Solve the resulting equation for the variable to find the critical points.

For the function given \(f(t) = t^2 - \frac{16}{t}\), we derived \(f'(t) = 2t + 16t^{-2}\). Setting this equal to zero gives us the critical points. Solving \(2t + \frac{16}{t^2} = 0\) eventually leads to \(t = 2\) when simplified.
Understanding the nature of these points requires us to carry out further analysis, such as using the second derivative test to conclude whether these points are maxima or minima.

Calculus

Calculus is the mathematics of change and is fundamental in many areas of science and engineering. It includes the study of limits, derivatives, integrals, and infinite series. When dealing with functions, calculus allows us to understand and interpret how functions behave.One of the core components of calculus is differentiation, which tells us how functions change with respect to variables—for example, calculating velocity as the rate of change of position. This differentiation leads us to concepts such as:

• First Derivative: Provides information on the slope of a function at any point. It helps in determining where the function has increasing or decreasing tendencies.
• Second Derivative: Offers insights into the function's concavity and helps clarify the nature of critical points as explored above.

Through techniques in calculus such as the power rule, chain rule, and product rule, we simplify the process of finding these derivatives, as was done when differentiating our function \(f(t) = t^2 - \frac{16}{t}\). With these derivatives, important properties like critical points and relative extrema are explored further.