Problem 29 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the absolute extrema of the function on the closed interval. Use a graphing utility to verify your results. $$ h(t)=(t-1)^{2 / 3}, \quad[-7,2] $$

Step-by-Step Solution

Answer

The absolute minimum of the function is 0 at \( t = 1 \) and the absolute maximum is 64 at \( t = -7 \).

Step 1: Identify the derivative of the function

We must first find the derivative \( h'(t) \) of the given function \( h(t) = (t-1)^{2 / 3} \). Using the rule \((u^n)' = n \cdot u^{n-1} \cdot u'\), we set \( u= (t-1) \) and \( n =\frac{2}{3} \). Hence, \( h'(t) = \frac{2}{3}(t-1)^{-1/3}\cdot 1\).

Step 2: Find the critical points

The critical points are values of \( t \) where \( h'(t)=0 \) or \( h'(t)\) does not exist. After checking, \( h'(t) \) never equals 0. However it does not exist when \( t=1 \). So, \( t=1 \) is a critical point.

Step 3: Evaluate the function at critical points and endpoints

Evaluate the function \( h(t) \) at the critical point \( t=1 \) and at the endpoints \( t=-7 \) and \( t=2 \). Hence, we determine \( h(1) = (1-1)^{2/3} = 0, h(-7) = (-7-1)^{2/3} = 64, h(2) = (2-1)^{2/3} = 1 \).

Step 4: Determine the absolute extrema

Finally, compare the function values from step 3. The smallest value is 0 at \( t=1 \), thus the absolute minimum is \( 0 \) at \( t=1 \). The largest value is \( 64 \) at \( t=-7 \). So, the absolute maximum is \( 64 \) at \( t=-7 \).

Step 5: Use a Graphing Utility to Verify the Results

Use a graphing utility to graph the function \( h(t) \) on the interval \([-7,2]\). Check if the absolute maximum and minimum occur at the points \( t=-7 \) and \( t=1 \), respectively which should confirm our earlier computation.

Key Concepts

Critical Points

When working with functions, finding critical points is crucial for determining where the function experiences changes in its behavior. A critical point is a point on the graph of a function where the derivative is zero or undefined. This is typically where a function's slope becomes horizontal (zero derivative) or where it may have a sharp corner or cusp (undefined derivative).

In our example, for the function \( h(t) = (t - 1)^{2/3} \), we calculated its derivative as \( h'(t) = \frac{2}{3} (t-1)^{-1/3} \). Notice, the derivative is never zero. However, it is undefined at \( t=1 \) because you cannot divide by zero. Thus, \( t=1 \) is a critical point.

Identifying these points helps us evaluate the function's behavior and determine its extrema, which are the highest or lowest points over the specified interval.

Derivative

A derivative is a mathematical tool that helps us understand how a function changes at any given point. It represents the slope of a function's graph at a particular spot. In simple terms, the derivative gives us the rate of change or the steepness of the curve.

To find the derivative of the function \( h(t) = (t - 1)^{2/3} \), we used the power rule from calculus. We set \( u = t - 1 \) and \( n = \frac{2}{3} \). The derivative rule \( (u^n)' = n \, \cdot \, u^{n-1} \, \cdot \, u' \) is then applied, yielding \( h'(t) = \frac{2}{3} (t-1)^{-1/3} \). Observe how the negative exponent indicates that this derivative becomes undefined at \( t=1 \), which corresponds to our critical point.

Understanding the derivative helps us plot the behavior of the function and pinpoint where the function reaches significant maxima or minima.

Closed Interval

A closed interval in mathematics is one where all the boundary points are included. It is depicted by the notation \( [a, b] \), where \( a \) and \( b \) are the endpoints, explicitly included in the interval.

For our exercise, the function \( h(t) \) is assessed over the closed interval \([-7, 2]\). Analyzing the function over this interval involves evaluating it at both the endpoints and any critical points that lie within this range. In this example, the critical point \( t=1 \) was specifically examined alongside endpoints \( t=-7 \) and \( t=2 \).

Evaluating functions over closed intervals helps ensure that we have adequately captured all possible absolute extrema, as critical behavior often occurs at these boundary points.

Graphing Utility

A graphing utility is a helpful tool that allows us to visually examine and analyze functions. It can range from a physical calculator to a software application, enabling users to create precise graphs by inputting mathematical functions.

Using a graphing utility to graph the function \( h(t) \) over the interval \([-7, 2]\) allows us to visually confirm our calculations. Through this tool, we can easily see that the maximum point occurs at \( t=-7 \) with a value of \( 64 \), and the minimum occurs at \( t=1 \) with a value of \( 0 \).

Visual verification provides significant reassurance of correctness and allows us to explore the function's behavior intuitively. It serves as a valuable step in confirming theoretical solutions in calculus.