Problem 45 - Calculus Early Transcendentals

Question

a. Locate the critical points of \(f\) b. Use the First Derivative Test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist). $$f(x)=x^{2 / 3}(x-5) \text { on }[-5,5]$$

Step-by-Step Solution

Answer

Answer: The absolute maximum and minimum values of the function \(f(x) = x^{\frac{2}{3}}(x-5)\) on the interval \([-5, 5]\) are both 0, occurring at \(x=-5\), \(x=0\), and \(x=5\).

Step 1: Find the first derivative of the function f(x)

To find the derivative, use the product rule (uv)'=u'v + uv' where \(u(x) = x^{\frac{2}{3}}\) and \(v(x) = x - 5\). First, compute the derivatives of u(x) and v(x): $$u'(x) = \frac{d}{dx}(x^{\frac{2}{3}}) = \frac{2}{3}x^{-\frac{1}{3}}$$ $$v'(x) = \frac{d}{dx}(x - 5) = 1$$ Now, apply product rule: $$f'(x) = u'(x)v(x) + u(x)v'(x) = \frac{2}{3}x^{-\frac{1}{3}}(x - 5) + x^{\frac{2}{3}}(1)$$

Step 2: Determine critical points of the function f(x)

A critical point exists where the derivative is 0 or does not exist. Solve for x in the equation \(f'(x) = 0\): $$\frac{2}{3}x^{-\frac{1}{3}}(x - 5) + x^{\frac{2}{3}}(1) = 0$$ Let \(w=x^{\frac{1}{3}}\), then the equation becomes quadratic in w: $$\frac{2}{3}w^2(w - 5) + w^3 = 0$$ $$2w^3 - 10w^2 + 3w^3 = 0$$ $$w^2(5w - 10) = 0$$ Now, we can find the values of x: $$w^2 = 0 \Rightarrow x = w^3 = 0$$ $$5w - 10 = 0 \Rightarrow w = 2 \Rightarrow x = w^3 = 8$$ So, the critical points are \(x = 0\) and \(x = 8\). However, as the interval is \([-5, 5]\) we will consider only \(x=0\).

Step 3: Use the First Derivative Test

Let's analyze whether the critical point \(x = 0\) is a local maximum or minimum using the signs of the derivative: $$f'(-1) = \frac{2}{3}(-1)^{-\frac{1}{3}}(-1 - 5) + (-1)^{\frac{2}{3}}(1) > 0$$ $$f'(1) = \frac{2}{3}(1)^{-\frac{1}{3}}(1 - 5) + (1)^{\frac{2}{3}}(1) < 0$$ Since the sign of the first derivative changes from positive to negative, the critical point at \(x = 0\) corresponds to a local maximum.

Step 4: Find absolute maximum and minimum

Now let's find the function value at the critical point and both endpoints of the interval \([-5, 5]\): $$f(-5) = (-5)^{\frac{2}{3}}(-5-5) = 0$$ $$f(0) = 0^{\frac{2}{3}}(0-5) = 0$$ $$f(5) = 5^{\frac{2}{3}}(5-5) = 0$$ The function values at all critical points and interval endpoints are equal to 0. Thus, we have an absolute maximum and minimum value of 0 at \(x=-5\), \(x=0\), and \(x=5\).

Key Concepts

First Derivative Test

When you are dealing with functions and their graphs, you often want to find the high and low points of the function. These are called local maximums and minimums. The First Derivative Test is a useful tool for this. It helps you determine whether a critical point is a local maximum, minimum, or neither. Here's how it works:

The first derivative of a function, denoted as \(f'(x)\), tells us the slope of the function at any given point. When \(f'(x) = 0\) or does not exist, these points are called critical points. To apply the First Derivative Test:

• Find the critical points by setting \(f'(x) = 0\) or finding where it is undefined.
• Determine the sign of \(f'(x)\) on either side of the critical point.
• If \(f'(x)\) changes from positive to negative at a critical point, it's a local maximum. If it changes from negative to positive, it's a local minimum.
• If there’s no sign change, the point is neither a maximum nor a minimum.

In the exercise, we identified \(x = 0\) as a critical point. Checking the signs to the left and right of \(x = 0\) showed a change from positive to negative, indicating a local maximum at this point.

Absolute Maximum and Minimum

Finding the absolute maximum and minimum values of a function is about identifying the highest and lowest points over a specified interval. For a continuous function on a closed interval, these values will occur either at critical points within the interval or at the endpoints of the interval.

To find the absolute maximum and minimum:

• Evaluate the function at its critical points.
• Evaluate the function at the endpoints of the interval.
• Compare these values to identify the largest (absolute maximum) and smallest (absolute minimum).

In the given problem, the function \(f(x) = x^{2/3}(x-5)\) was examined over the interval \([-5, 5]\). The evaluations revealed that the critical point \(x = 0\) and the endpoints \(x = -5\) and \(x = 5\) all returned a function value of 0. Therefore, the absolute maximum and minimum values on this interval are both 0, occurring at these points.

Product Rule Derivatives

When you have two functions multiplied together, like \(u(x)\) and \(v(x)\), and you need to take the derivative of their product \(u(x)v(x)\), you'll use the Product Rule. This is a fundamental tool in calculus and is stated as follows: \[(uv)' = u'v + uv'\]Here's how it works in practice:

• First, differentiate each of the functions separately to find \(u'(x)\) and \(v'(x)\).
• Use the formula to plug these derivatives into \((uv)' = u'v + uv'\).
• Simplify to find the derivative of the original product function.

In the exercise, the function \(f(x) = x^{2/3}(x-5)\) required us to differentiate the product of \(x^{2/3}\) (as \(u(x)\)) and \(x - 5\) (as \(v(x)\)). Using the Product Rule, we found the derivative \(f'(x) = \frac{2}{3}x^{-rac{1}{3}}(x - 5) + x^{\frac{2}{3}}(1)\). This derivative was key in determining the critical points and analyzing changes in the function.