Problem 31 - Calculus with Concepts in Calculus

Question

Use the Second Derivative Test to determine the relative extreme values (if any) of the function. $$ f(t)=t^{2}+1+1 / t $$

Step-by-Step Solution

Answer

The function has a relative minimum at \( t = \sqrt[3]{\frac{1}{2}} \).

Step 1: Find the First Derivative

Start by finding the first derivative of the given function, \( f(t) = t^2 + 1 + \frac{1}{t} \). Using standard differentiation rules, \( \frac{d}{dt}(t^2) = 2t \) and \( \frac{d}{dt}(\frac{1}{t}) = -\frac{1}{t^2} \). So, the first derivative is \( f'(t) = 2t - \frac{1}{t^2} \).

Step 2: Set the First Derivative to Zero

To find critical points, set \( f'(t) = 0 \). The equation becomes \( 2t - \frac{1}{t^2} = 0 \). Multiply everything by \( t^2 \) to clear the fraction, resulting in \( 2t^3 - 1 = 0 \). Thus, \( 2t^3 = 1 \) or \( t^3 = \frac{1}{2} \), so \( t = \sqrt[3]{\frac{1}{2}} \).

Step 3: Find the Second Derivative

Next, find the second derivative from \( f'(t) = 2t - \frac{1}{t^2} \). Differentiating, \( f''(t) = 2 + \frac{2}{t^3} \) because \( \frac{d}{dt}(-\frac{1}{t^2}) = \frac{2}{t^3} \).

Step 4: Apply the Second Derivative Test

To determine the nature of the critical point \( t = \sqrt[3]{\frac{1}{2}} \), evaluate the second derivative there. Substitute \( t = \sqrt[3]{\frac{1}{2}} \) in \( f''(t) \): \( f''(\sqrt[3]{\frac{1}{2}}) = 2 + \frac{2}{\left(\sqrt[3]{\frac{1}{2}}\right)^3} = 2 + 4 = 6 \). Since \( f''(t) > 0 \), the function has a relative minimum at \( t = \sqrt[3]{\frac{1}{2}} \).

Key Concepts

First Derivative

The first derivative of a function, denoted as \( f'(t) \), is a crucial concept in calculus that represents the slope of the tangent line at any point on the function's graph. For the given function \( f(t) = t^2 + 1 + \frac{1}{t} \), finding the first derivative involves differentiating each term individually.

• The derivative of \( t^2 \) is \( 2t \).
• The derivative of the constant 1 is 0, as constants do not change.
• The derivative of \( \frac{1}{t} \) is \( -\frac{1}{t^2} \), achieved by applying the power rule for negative exponents.

Combining these results provides the first derivative: \( f'(t) = 2t - \frac{1}{t^2} \). This equation is essential for finding critical points, which help us identify where the function might experience extreme values.

Critical Points

In the context of calculus, critical points are where potential relative minima or maxima occur. To find them, set the first derivative \( f'(t) \) to zero and solve for \( t \). For this exercise, we're dealing with the equation:

• \( 2t - \frac{1}{t^2} = 0 \)

Simplifying this by multiplying through by \( t^2 \) yields \( 2t^3 - 1 = 0 \). Further simplification gives \( t^3 = \frac{1}{2} \), resulting in \( t = \sqrt[3]{\frac{1}{2}} \). This value of \( t \) represents our critical point. Identifying it is essential because it indicates where the function's slope is zero. However, determining whether this point is a relative minimum or maximum requires additional analysis of the second derivative.

Second Derivative

The second derivative, \( f''(t) \), gives insights into the function's concavity and helps confirm if a critical point is a minimum or maximum. To find the second derivative, differentiate the first derivative \( f'(t) = 2t - \frac{1}{t^2} \):

• The derivative of \( 2t \) is 2.
• The derivative of \( -\frac{1}{t^2} \) is \( \frac{2}{t^3} \), through careful application of the power rule for negative exponents.

Thus, the second derivative is \( f''(t) = 2 + \frac{2}{t^3} \). This expression is vital for applying the Second Derivative Test, which assists in determining the nature of critical points.

Relative Extreme Values

Relative extreme values are points where a function attains a local minimum or maximum. The Second Derivative Test helps identify these values at critical points. To apply the test, evaluate the second derivative \( f''(t) \) at the critical point. For this exercise, the critical point is \( t = \sqrt[3]{\frac{1}{2}} \).

• Substitute this into \( f''(t) = 2 + \frac{2}{t^3} \).
• Calculate: \( f''(\sqrt[3]{\frac{1}{2}}) = 2 + 4 = 6 \).

The result, 6, is greater than zero, indicating a positive concavity at this point. Therefore, the function has a relative minimum at \( t = \sqrt[3]{\frac{1}{2}} \). This conclusion is pivotal, as it confirms not just the presence but the nature of relative extreme values within the function's behavior.