Problem 51 - Calculus Early Transcendentals

Question

Show that the absolute minimum value of $$f(x)=x^{2}+\frac{x^{2}}{(8-x)^{2}}, \quad x>8$$ occurs at \(x=10\) by using a CAS to find \(f^{\prime}(x)\) and to solve the equation \(f^{\prime}(x)=0\)

Step-by-Step Solution

Answer

The absolute minimum value of the function occurs at \(x=10\).

Step 1: Calculate the derivative

First, we need to find the derivative of the function \( f(x) = x^{2} + \frac{x^{2}}{(8-x)^{2}} \) with respect to \( x \). We'll use standard differentiation techniques. Let's differentiate both terms individually:1. The derivative of \( x^2 \) is \( 2x \).2. Use the quotient rule for the second term, \( \frac{x^2}{(8-x)^2} \): - Let \( u = x^2 \) and \( v = (8-x)^2 \). - \( u' = 2x \). - \( v' = -2(8-x) \). The quotient rule is \( \frac{u'v - uv'}{v^2} \), so the derivative becomes: \[ \left( \frac{x^2}{(8-x)^2} \right)' = \frac{2x(8-x)^2 - x^2(-2)(8-x)}{(8-x)^4} = \frac{2x(8-x)^2 + 2x^2(8-x)}{(8-x)^4} \].

Step 2: Simplify the derivative expression

Now, simplify the expression we obtained from the derivative of the second term:\[ \frac{2x(8-x)^2 + 2x^2(8-x)}{(8-x)^4} \] becomes: - Since both terms in the numerator contain \(2x\), factor out \(2x\): \[ 2x[(8-x)^2 + x(8-x)] = 2x[64 - 16x + x^2 + 8x - x^2] \] Simplify inside the brackets: \[ 2x[64 - 8x] = 128x - 16x^2 \].Thus, the derivative can be written as:\( f'(x) = 2x + \frac{128x - 16x^2}{(8-x)^4} \).

Step 3: Set the derivative equal to zero

We set the derivative \( f'(x) \) equal to zero to find the critical points.\[ 2x + \frac{128x - 16x^2}{(8-x)^4} = 0 \]Multiply through by \((8-x)^4\) to eliminate the fraction:\[ 2x(8-x)^4 + (128x - 16x^2) = 0 \].

Step 4: Solve the equation for critical points

Using a computer algebra system (CAS), solve the equation:\[ 2x(8-x)^4 + 128x - 16x^2 = 0 \]The solution found by the CAS shows that \( x = 10 \) is a critical point.

Step 5: Verify if the critical point is a minimum

To verify that \( x = 10 \) is a minimum, check the second derivative, \( f''(x) \), or analyze the behavior of \( f'(x) \) around \( x = 10 \) (e.g., point decreases and increases).For simplicity, let's recognize that in a given interval, \( f(x) \rightarrow \infty \) as \( x \rightarrow 8^+ \) or \( x \rightarrow \infty \), thus showing \( x = 10 \) as a minimum.The test or plotting \( f'(x) \) confirms that \( f'(x) \) changes from negative to positive at \( x = 10 \), which means \( x = 10 \) is a minimum point.

Key Concepts

Derivative

When exploring calculus, particularly within differentiation, the derivative is a fundamental concept. The derivative represents the rate at which a function is changing at any given point. It provides insight into the slope of the function's graph at that specific point. In mathematical terms, if we have a function denoted by \( f(x) \), its derivative \( f'(x) \) describes how \( f(x) \) changes as \( x \) varies.
For example, calculating the derivative of the function \( f(x) = x^2 + \frac{x^2}{(8-x)^2} \) involves differentiating each term separately. The derivative of \( x^2 \) is straightforward, given by \( 2x \). The second term is more complex and requires the application of specific rules, like the quotient rule, to obtain its derivative.

Critical Points

Critical points are crucial in understanding the behavior of functions. These are the values of \( x \) where the derivative of a function \( f'(x) \) is zero or undefined. Critical points help identify potential local maxima or minima of a function.
In our exercise, finding the critical points involves solving the equation \( f'(x) = 0 \). After substituting the expression for the derivative, critical points are determined using methods such as setting the resulting equation equal to zero and solving for \( x \).
In the scenario described, by employing a CAS or manual calculation, it turns out that \( x = 10 \) is a critical point, highlighting a location where the behavior of the function may change.

Quotient Rule

The quotient rule is an essential tool in calculus for differentiating functions that are in the form of one function divided by another. When encountering a function \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \), the derivative is found using:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
To differentiate \( f(x) = \frac{x^2}{(8-x)^2} \), let \( u = x^2 \) and \( v = (8-x)^2 \). Compute \( u' \) and \( v' \), and apply the quotient rule to find the derivative of this term. This yields an expression that must be further simplified for use in finding critical points.

Second Derivative Test

The second derivative test is a method used to determine the nature of critical points found in a function. By analyzing \( f''(x) \), the second derivative of \( f(x) \), one can infer whether a critical point is a local minimum, local maximum, or neither:

• If \( f''(x) > 0 \) at a critical point, the function is concave up, indicating a local minimum.
• If \( f''(x) < 0 \), the function is concave down, pointing to a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and further analysis or graphing is required.

In the context of the exercise, although the detailed second derivative wasn't calculated, it's important to note that the behavior of \( f'(x) \) around \( x = 10 \) confirms that it shifts from negative to positive, indicating a local minimum at this point.