Problem 22 - Calculus with Concepts in Calculus

Question

Find all extreme values (if any) of the given function on the given interval. Determine at which numbers in the interval these values occur. $$ f(t)=t^{2}-4 / t ;[1,3) $$

Step-by-Step Solution

Answer

Minimum value is -3 at \( t = 1 \); maximum approached as \( t \rightarrow 3^- \) is 7.67.

Step 1: Find the derivative of the function

The first step is to find the derivative of the function \( f(t) = t^2 - \frac{4}{t} \). Use the power rule and the derivative of \( \frac{1}{t} \) to find \( f'(t) \). This results in \[ f'(t) = 2t + \frac{4}{t^2}. \]

Step 2: Set the derivative equal to zero

To find the critical points, set the derivative equal to zero: \( 2t + \frac{4}{t^2} = 0 \). Multiply through by \( t^2 \) to eliminate the fraction, resulting in \( 2t^3 + 4 = 0 \).

Step 3: Solve the equation for critical points

Solve the equation \( 2t^3 + 4 = 0 \) by isolating \( t^3 \): \( 2t^3 = -4 \). Divide by 2: \( t^3 = -2 \). Take the cube root: \( t = -\sqrt[3]{2} \). Since this is not in the interval \([1,3)\), it is not a critical point for our interval.

Step 4: Check endpoints of the interval

Evaluate the function at the endpoints of the interval (the interval is closed at 1 and open at 3). Compute \( f(1) \) and check the limit as \( t \rightarrow 3^- \). Hence: \[ f(1) = 1^2 - \frac{4}{1} = 1 - 4 = -3, \] and the limit as \( t \rightarrow 3^- \) is \[ \lim_{t \to 3^-} (t^2 - \frac{4}{t}) = 9 - \frac{4}{3} = 9 - 1.33 = 7.67. \]

Step 5: Determine extreme values

Since there are no critical points in the interval \([1,3)\), check only the function's values at the endpoint \(t=1\) and as \(t\) approaches 3 from the left. Compare these values. Since \( f(1) = -3 \) and \( \lim_{t \to 3^-} f(t) = 7.67 \), \( f(1) = -3 \) is the minimum value, and 7.67 is the maximum value approached as \( t \) approaches 3 from the left.

Key Concepts

Critical Points

In calculus, critical points are special places on a function's graph where the gradient (or slope) is zero or undefined. These points often correspond to local maximums or minimums. To find critical points, we derive the function to get the derivative, and then solve for when this derivative is equal to zero.

For the function \( f(t) = t^2 - \frac{4}{t} \), we calculated the derivative as \( f'(t) = 2t + \frac{4}{t^2} \). Setting this derivative to zero, we solved:

• \( 2t + \frac{4}{t^2} = 0 \)
• Multiply by \(t^2\) to clear the fraction, giving \( 2t^3 + 4 = 0 \)
• Solving this, we found \( t = -\sqrt[3]{2} \)

However, this critical point does not lie within the interval \([1, 3)\), hence it is irrelevant for determining extreme values on this specific interval.

Derivative

Derivatives are mathematical tools that describe the rate of change of a function. They are essential for finding critical points and thus determining extreme values. By taking the derivative of a function, we can analyze its behavior regarding when it increases or decreases.

In this exercise, we used derivative rules:

• For power functions like \( t^2 \), the derivative is found using the power rule resulting in \( 2t \)
• For fractions like \( \frac{4}{t} \), the derivative is \( -\frac{4}{t^2} \) because \( \frac{1}{t} \) differentiates to \( -\frac{1}{t^2} \)

The result, \( f'(t) = 2t + \frac{4}{t^2} \), helped identify where the slope of the function becomes zero, a key step in finding critical points.

Interval Notation

Interval notation is a concise way to describe a portion of the real number line. It is read as elements that lie within or are around specified bounds but do not necessarily meet them fully.

In this problem, we examined the interval \([1, 3)\). This means:

• The interval includes every point between 1 and 3
• It is closed at 1, meaning 1 is included in the interval
• It is open at 3, meaning 3 is not included

Understanding this notation is crucial for determining where to evaluate the function, as it defines the "valid" zone where we search for maximums or minimums.

Limit Evaluation

Limit evaluation is used to understand the behavior of functions as they approach a specific point, particularly when that point isn't directly reachable within a given interval.

For this function, \( f(t) = t^2 - \frac{4}{t} \), evaluating limits was necessary at the open endpoint. We computed:

• At \( t = 1 \), we directly plugged the value into the function, finding \( f(1) = -3 \)
• For \( t \) approaching 3 from the left, we calculated: \( \lim_{t \to 3^-} (t^2 - \frac{4}{t}) = 7.67 \)

This analysis showed the function reaches its maximum at an approach to 3 and a minimum at the endpoint 1 within the interval. Limits are powerful in assessing boundary behavior even when full evaluation is impossible.