Problem 18 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Consider the function \(f(x)=x^{5}-2 x^{4}-7\) restricted to the domain \([-1,1]\). Your reasoning for the questions below must be fully explained and be independent of a graphing calculator. (a) Find the absolute maximum value of \(f(x)\) on the interval \([-1,1]\) or explain why this is not possible. (b) Find the absolute minimum value of \(f(x)\) on the interval \([-1,1]\) or explain why this is not possible. (c) Find the absolute minimum value of \(f(x)\) on the open interval \((-1,1)\) or explain why this is not possible.

Step-by-Step Solution

Answer

The absolute maximum value of \(f(x)\) on the interval \([-1,1]\) is -7 at \(x = 0\) and the absolute minimum is -10 at \(x = -1\). The absolute minimum value of \(f(x)\) on the open interval \((-1, 1)\) is -7 at \(x = 0\).

Step 1: Find the derivative

Calculate the derivative of the function \(f(x) = x^{5} - 2x^{4} - 7\). The derivative is calculated as \(f'(x) = 5x^{4} - 8x^{3}\).

Step 2: Find the critical points

Set the derivative equal to zero and solve for x to find critical points. This gives us \(5x^{4} - 8x^{3} = 0\). Factor out \(x^{3}(5x - 8) = 0\), so possible critical points are \(x = 0\) or \(x = 8/5\) but only \(x = 0\) lies in the interval \([-1, 1]\).

Step 3: Evaluate the function

Evaluate the function at the critical point and at the endpoints of the interval. Thus, evaluate \(f(x)\) at \(x = -1\), \(x = 0\) and \(x = 1\). This gives us \(f(-1) = -10\), \(f(0) = -7\), and \(f(1) = -8\).

Step 4: Determine the maximum and minimum

Compare the values of \(f(x)\) at the critical point and at the endpoints. The absolute maximum of \(f(x)\) on the interval \([-1,1]\) is -7 at \(x = 0\) and the absolute minimum is -10 at \(x = -1\).

Step 5: Consider the open interval

For the open interval, only the critical point and interior values matter. Since we only have one critical point within this interval, the absolute minimum value of \(f(x)\) on the open interval \((-1, 1)\) is again -7 at \(x = 0\). The endpoints aren't considered because it's an open interval.

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. It's a fundamental tool to determine various characteristics of functions. For the function provided, \( f(x) = x^5 - 2x^4 - 7 \), the first step is to find the derivative \( f'(x) \). This involves applying basic rules of differentiation. The power rule is particularly useful here, which states that if you have a term \( ax^n \), its derivative will be \( nax^{n-1} \). For our function:

• First term: \( 5x^5 \) becomes \( 5 \times 1x^{5-1} = 5x^4 \)
• Second term: \(-2x^4\) becomes \(-8x^{4-1} = -8x^3 \).
• Constant term: \(-7\) has a derivative of 0, as it does not change with x.

Thus, the derivative is \( f'(x) = 5x^4 - 8x^3 \). Finding the derivative helps us examine how the function's output changes over an interval and identify points where it doesn’t change at all – the critical points.

Critical Points

Critical points of a function are where the derivative is zero or undefined. These points can indicate where a function may have a maximum, minimum, or a point of inflection. For the function in this exercise, setting the derivative \( f'(x) = 5x^4 - 8x^3 \) to zero helps us find these critical points:

• Factor the expression: \( x^3(5x - 8) = 0 \).
• Solutions are \( x = 0 \) or \( x = \frac{8}{5} \).

However, the exercise limits us to the interval \([-1,1]\). Hence, the only critical point that fits within this interval is \( x = 0 \). Critical points give us valuable insight into where the function's rate of change pauses, which is key to determining maxima and minima.

Maximum and Minimum Values

To find the absolute maximum and minimum values of a function within a specific interval, we evaluate the function at critical points and endpoints of that interval. Respective function values are compared.For the function \( f(x) = x^5 - 2x^4 - 7 \) on \([-1, 1]\), evaluate it at:

• Endpoints: \( f(-1) = -10 \) and \( f(1) = -8 \).
• Critical Point: \( f(0) = -7 \).

Comparing these values, the absolute minimum is \(-10\) at \( x = -1 \) and the absolute maximum is \(-7\) at \( x = 0 \).Maximum and minimum values provide insights into the highest and lowest points a function attains within a given range. These concepts are especially important in optimization problems where understanding the range of values a function can take is crucial.

Closed and Open Intervals

Intervals in calculus can be closed or open, and they dictate how we evaluate functions within those intervals. A closed interval \([a, b]\) includes its endpoints, while an open interval \((a, b)\) does not.In part (a) and (b) of the exercise, we deal with a closed interval \([-1, 1]\). This means evaluating \( f(x) \) at its endpoints and any critical points within the interval. On closed intervals, maximum and minimum values can occur at endpoints or critical points.Part (c) considers an open interval \((-1, 1)\). In this case, endpoints like \( x = -1, \text{and} x = 1\) are not included. We only consider critical points within this boundary. Here, since the critical point \( x = 0 \) falls inside, \( -7 \) remains the absolute minimum within this open interval.Understanding these differences helps in analyzing functions accurately and is particularly crucial when dealing with optimization and determining function behavior in specific ranges.