Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ h(t)=\frac{t^{5 / 3}}{2+t} ; I=[-1,8] $$

Step-by-Step Solution

Answer

The maximum value is approximately 17.86 at t = 5, and the minimum value is -1 at t = -1.

Step 1: Find the derivative of the function

To find the critical points of the function \( h(t) = \frac{t^{5/3}}{2+t} \), we first need to calculate its derivative. We use the quotient rule for derivatives, which is given by: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]. Here, \( u = t^{5/3} \) and \( v = 2 + t \). Thus, \( u' = \frac{5}{3}t^{2/3} \) and \( v' = 1 \). The derivative \( h'(t) \) is: \[ h'(t) = \frac{\frac{5}{3} t^{2/3}(2+t) - t^{5/3} \cdot 1}{(2+t)^2} \].

Step 2: Simplify the derivative expression

Simplify the expression for \( h'(t) \): \[ h'(t) = \frac{(\frac{5}{3}t^{2/3})(2+t) - t^{5/3}}{(2+t)^2} = \frac{\frac{10}{3}t^{2/3} + \frac{5}{3}t^{5/3} - t^{5/3}}{(2+t)^2} \]. Combine like terms: \[ h'(t) = \frac{\frac{10}{3}t^{2/3} - \frac{2}{3}t^{5/3}}{(2+t)^2} \].

Step 3: Set the derivative to zero and solve

To find critical points, set \( h'(t) = 0 \). This means the numerator should be zero: \( \frac{10}{3}t^{2/3} - \frac{2}{3}t^{5/3} = 0 \). Factor out \( \frac{2}{3}t^{2/3} \): \[ \frac{2}{3}t^{2/3}(5 - t) = 0 \]. Thus, \( t = 0 \) or \( t = 5 \) are critical points.

Step 4: Evaluate the function at critical points and endpoints

Find \( h(t) \) at critical points and endpoints of the interval \( [-1, 8] \). Calculate \( h(-1) = \frac{(-1)^{5/3}}{1} = -1 \), \( h(0) = \frac{0^{5/3}}{2} = 0 \), \( h(5) = \frac{5^{5/3}}{7} = \frac{125}{7} \), and \( h(8) = \frac{8^{5/3}}{10} = \frac{32}{5} \).

Step 5: Determine the maximum and minimum values

Compare \( h(t) \) at all calculated points: \( h(-1) = -1 \), \( h(0) = 0 \), \( h(5) = \frac{125}{7} \approx 17.86 \), \( h(8) = \frac{32}{5} = 6.4 \). The minimum value is \(-1\) at \( t = -1 \), and the maximum value is approximately \( 17.86 \) at \( t = 5 \).

Key Concepts

Quotient Rule

In calculus, the Quotient Rule is a fundamental tool used to differentiate functions that are represented as a quotient, meaning one function divided by another. When you have a function like \( h(t) = \frac{u(t)}{v(t)} \), the Quotient Rule helps us find the derivative of this function, denoted \( h'(t) \). It states:

• \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)

To apply the Quotient Rule, you need to:

• Identify the numerator \( u(t) \) and the denominator \( v(t) \) of your function.
• Differentiate these individually to get \( u'(t) \) and \( v'(t) \).
• Substitute these derivatives into the Quotient Rule formula.
• Simplify the result to get the derivative \( h'(t) \).

Using this rule requires caution because it involves multiple steps, and errors can occur easily if each derivative is not computed with care. For example, in \( h(t) = \frac{t^{5/3}}{2+t} \), \( u = t^{5/3} \) and \( v = 2 + t \), where \( u' = \frac{5}{3}t^{2/3} \) and \( v' = 1 \). Applying the Quotient Rule gives us the derivative expression before simplification.

Derivative Simplification

After finding the derivative of a function, in many cases, it will usually appear complicated at first. Derivative simplification is an important step in calculus that helps in making the expression easier to work with, especially for subsequent calculations like finding critical points.The primary goal of simplifying a derivative is to reduce the complexity of the expression, often by combining like terms or factoring when possible. For the given solution involving \( h'(t) \), simplification was crucial because the initial derivative \( h'(t) = \frac{\frac{5}{3} t^{2/3}(2+t) - t^{5/3}}{(2+t)^2} \) looks quite complex.To simplify it, you combine terms in the numerator. By carefully distributing and factoring out common terms such as \( \frac{2}{3}t^{2/3} \), you can then transform it to \( \frac{\frac{10}{3}t^{2/3} - \frac{2}{3}t^{5/3}}{(2+t)^2} \). This simpler form is much easier to use for setting it equal to zero and finding critical points.

Maximum and Minimum Values

Finding maximum and minimum values of a function within a closed interval involves checking the function at critical points and at the interval's endpoints. Critical points arise where the derivative either equals zero or does not exist. These points are significant because they can indicate where the function changes direction, thus potentially being a maximum or minimum.Once you identify the critical points, like \( t = 0 \) and \( t = 5 \) found through setting the derivative to zero, evaluate the function values at these points along with the endpoints of the interval \([-1, 8]\). This step ensures you thoroughly check all potential extremes of the function. Evaluating \( h(t) \) at \(-1, 0, 5,\) and \(8\) provides the values \(-1, 0, \frac{125}{7},\) and \(\frac{32}{5}\), respectively.By comparing these values, you determine the minimum and maximum values over the interval. In this example, the minimum is \(-1\) at \( t = -1 \) and the maximum is approximately \( 17.86 \) at \( t = 5 \).

Interval Evaluation

Evaluating a function over a specified interval is essential when analyzing the behavior of the function within certain limits. It involves checking the behavior at the boundaries as well as any critical points found inside the interval.For the given function and interval \([-1, 8]\), it means we must inspect how \( h(t) \) behaves from \( t = -1 \) to \( t = 8 \). By calculating \( h(t) \) at the critical points and the interval's endpoints, we get all potential maximum and minimum values within this range.Remember that sometimes functions can have interesting behavior at the edges of an interval, and neglecting these points could lead to overlooking the actual extremes. This comprehensive evaluation provides a full picture of how the function behaves over the interval. In intervals like \([-1, 8]\), evaluating at \(-1, 0, 5,\) and \(8\) allowed us to conclude on the function's bounds within the intervals effectively.