Problem 36 - Calculus with Concepts in Calculus

Question

Use the Second Derivative Test to determine the relative extreme values (if any) of the function. $$ f(t)=t^{2}-8 \ln t $$

Step-by-Step Solution

Answer

The function has a local minimum at \( t = 2 \).

Step 1: Find the First Derivative

To apply the second derivative test, we first need to find the first derivative of the function \( f(t) = t^2 - 8 \ln t \). Differentiate each term with respect to \( t \):- The derivative of \( t^2 \) is \( 2t \).- The derivative of \(-8 \ln t \) is \( -\frac{8}{t} \).Thus, the first derivative is:\[ f'(t) = 2t - \frac{8}{t} \].

Step 2: Solve for Critical Points

To find critical points, set \( f'(t) = 0 \) and solve for \( t \):\[ 2t - \frac{8}{t} = 0 \]Multiply through by \( t \) (assuming \( t eq 0 \)) to clear the fraction:\[ 2t^2 = 8 \]Solve for \( t^2 \):\[ t^2 = 4 \]Taking the square root of both sides gives:\( t = 2 \) or \( t = -2 \). Since \( \ln t \) is undefined for \( t \leq 0 \), we only consider \( t = 2 \).

Step 3: Find the Second Derivative

Differentiate \( f'(t) = 2t - \frac{8}{t} \) to find the second derivative:- The derivative of \( 2t \) is \( 2 \).- The derivative of \(-\frac{8}{t} \) is \( \frac{8}{t^2} \).Thus, the second derivative is:\[ f''(t) = 2 + \frac{8}{t^2} \].

Step 4: Apply the Second Derivative Test at the Critical Point

Use \( f''(t) \) to determine the nature of the critical point \( t = 2 \):Substitute \( t = 2 \) into \( f''(t) \):\[ f''(2) = 2 + \frac{8}{2^2} = 2 + 2 = 4 \].Since \( f''(2) > 0 \), \( t = 2 \) is a local minimum.

Key Concepts

First Derivative

The first derivative of a function is a critical tool in calculus. It helps us understand how the function is changing at any point on its curve. When we talk about the first derivative, denoted as \( f'(t) \), we are essentially looking at the slope of the tangent line to the function at any given point. If this derivative is positive, the function is increasing; if it's negative, the function is decreasing.

In our exercise, to find the first derivative of \( f(t) = t^2 - 8 \ln t \), we differentiate each part of the function separately. The derivative of \( t^2 \) is \( 2t \), and the derivative of \( -8 \ln t \) is \( -\frac{8}{t} \). Putting these together, the first derivative becomes:

• \( f'(t) = 2t - \frac{8}{t} \)

Understanding this derivative helps us find critical points, which are important for identifying features like maximums and minimums in a function.

Critical Points

Critical points are where the function's first derivative is zero or undefined. These points are significant because they are potential locations where the function could have a relative maximum or minimum, or sometimes neither.

To find these points for our function, we set the first derivative \( f'(t) = 2t - \frac{8}{t} \) equal to zero:

• \( 2t - \frac{8}{t} = 0 \).

To solve this, we clear the fraction by multiplying through by \( t \), resulting in \( 2t^2 = 8 \). Solving for \( t^2 \) gives us \( t^2 = 4 \), leading to potential critical points: \( t = 2 \) and \( t = -2 \).

However, since the original function includes \( \ln t \), which is not defined for non-positive values, we only consider \( t = 2 \) as a valid critical point.

Second Derivative

The second derivative gives us information about the concavity of the function, which tells us if the curve is bending upwards or downwards at a particular point.

In this exercise, we compute the second derivative of the function using the first derivative expression \( f'(t) = 2t - \frac{8}{t} \). Differentiating this again, we get:

• The derivative of \( 2t \) is \( 2 \).
• The derivative of \( -\frac{8}{t} \) is \( \frac{8}{t^2} \).

Thus, the second derivative is:

• \( f''(t) = 2 + \frac{8}{t^2} \)

This expression helps us decide whether the function has a local maximum or minimum at a critical point.

Local Minimum

A local minimum is a point where the function value is lower than any other nearby points. To determine if a critical point is a local minimum or maximum, we can use the second derivative test.

For our function, the relevant critical point is \( t = 2 \). By substituting \( t = 2 \) into the second derivative \( f''(t) = 2 + \frac{8}{t^2} \), we perform the check:

• \( f''(2) = 2 + \frac{8}{4} = 4 \)

Since \( f''(2) > 0 \), the function is concave up at \( t = 2 \), indicating that it is a local minimum.

This means at \( t = 2 \), our function \( f(t) = t^2 - 8 \ln t \) has a valley-like point, where it's lower than at surrounding points.