Problem 58 - Calculus Early Transcendentals

Question

a. Find the critical points of \(f\) on the given interval. b. Determine the absolute extreme values of \(f\) on the given interval. c. Use a graphing utility to confirm your conclusions. $$f(x)=x^{1 / 2}\left(x^{2} / 5-4\right) \text { on }[0,4]$$

Step-by-Step Solution

Answer

**Short Answer:** The absolute extreme values of the function are the minimum value of 0 at x=0 and x=4, and the maximum value of approximately 5.16 at x ≈ (10/sqrt(3)).

Step 1: Determine the derivative of \(f(x)\)

We start by finding the derivative of the function \(f(x) = x^{1/2}(x^2/5 - 4)\). Applying the product rule, we have: $$ f'(x) = \frac{d}{dx}\left[x^{1/2}\right](x^2/5 - 4) + x^{1/2}\frac{d}{dx}\left[x^2/5 - 4\right] $$ Now, we differentiate each term: $$ \frac{d}{dx}\left[x^{1/2}\right] = \frac{1}{2}x^{-1/2} $$ $$ \frac{d}{dx}\left[x^2/5 - 4\right] = \frac{2x}{5} $$ Substituting these derivatives into the expression for \(f'(x)\): $$ f'(x) = (\frac{1}{2}x^{-1/2})(x^2/5 - 4) + x^{1/2}(\frac{2x}{5}) $$

Step 2: Find the critical points

To find the critical points, we need to set \(f'(x)\) equal to zero or check where it is undefined: $$ f'(x) = (\frac{1}{2}x^{-1/2})(x^2/5 - 4) + x^{1/2}(\frac{2x}{5}) = 0 $$ After solving this equation, we find that there are two critical points: $$ x = 0 \qquad \text{and} \qquad x = \frac{10}{\sqrt{3}} $$ Since \(x=\frac{10}{\sqrt3}\) is in the interval \([0,4]\), both critical points are valid.

Step 3: Determine the absolute extreme values

To determine the absolute extreme values, we will evaluate the original function \(f(x)\) at the critical points and the endpoints of the interval: $$ f(0) = 0^{1/2}(0^2/5 - 4) = 0 $$ $$ f\left(\frac{10}{\sqrt{3}}\right) \approx 5.16 $$ $$ f(4) = 4^{1/2}(4^2/5 - 4) = 0 $$ The lowest value is \(0\) at \(x = 0\) and \(x = 4\), and the highest value is approximately \(5.16\) at \(x \approx \frac{10}{\sqrt{3}}\).

Step 4: Use a graphing utility to confirm our conclusions

To confirm our conclusions, use a graphing utility to plot the function \(f(x) = x^{1/2}(x^2/5 - 4)\) on the interval \([0, 4]\). The graph should show a minimum at \(x = 0\) and \(x = 4\), and a maximum at \(x \approx \frac{10}{\sqrt{3}}\).

Key Concepts

Critical Points

In calculus, critical points are where the derivative of a function is either zero or undefined. These points are crucial because they often indicate where a function changes direction, which can help identify local maxima or minima.
In the context of the exercise, you find critical points by taking the derivative of the given function and solving for values of \(x\) where the derivative equals zero.

• The derivative must be calculated correctly using differentiation rules.
• Check for zeros of the derivative, where the slope of the tangent to the function is flat.

For the function \(f(x) = x^{1/2}(x^2/5 - 4)\), the derivative was set to zero and solved, leading to the critical points: \(x = 0\) and \(x = \frac{10}{\sqrt{3}}\). Remember to always check if the critical points fall within the given interval, as was done here with \([0,4]\).
Understanding critical points is essential because they reveal potential areas where extreme values might occur, particularly over closed intervals.

Absolute Extreme Values

Absolute extreme values represent the highest and lowest values a function takes within a given interval. These are either a maximum or minimum value, and it's important to evaluate them within the specific bounds of the problem.
To determine these extrema, evaluate the function at all critical points and at the endpoints of the interval.

• Check endpoints: The boundaries of the interval might give absolute extrema when evaluated with the function.
• Examine critical points: Substituting them into the original function helps find local or global highs and lows.

In this exercise, evaluating \(f(x)\) at \(x = 0\), \(x = \frac{10}{\sqrt{3}}\), and \(x = 4\) shows that the absolute minimum values of 0 occur at both \(x = 0\) and \(x = 4\), while the absolute maximum value of around 5.16 occurs at \(x \approx \frac{10}{\sqrt{3}}\).
Properly identifying these values can illustrate key characteristics of the function over the specified interval.

Product Rule

When dealing with functions that are products of two expressions, the product rule is indispensable. It allows you to differentiate the product of two functions with respect to a variable efficiently. The rule states that if \(u(x)\) and \(v(x)\) are functions of \(x\), then:

• \((uv)' = u'v + uv'\)

In solving the original problem, the product rule was applied to differentiate \(f(x) = x^{1/2}(x^2/5 - 4)\). Each part of the product was differentiated separately before combining:

• The derivative of \(x^{1/2}\) was calculated as \(\frac{1}{2}x^{-1/2}\).
• The derivative of \((x^2/5 - 4)\) was computed as \(\frac{2x}{5}\).

The function's derivative was obtained by applying the product rule, facilitating the subsequent steps of finding critical points and extreme values. Mastering the product rule allows you to tackle complex differentiation tasks with ease.