Problem 35 - Calculus Early Transcendentals

Question

a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing utility to determine whether each critical point corresponds to a local maximum, local minimum, or neither. $$f(x)=x^{2} \sqrt{x+1} \text { on }[-1,1]$$

Step-by-Step Solution

Answer

Question: Identify the critical points of the function f(x) = x^2 * sqrt(x+1) in the closed interval [-1, 1], and determine whether they correspond to a local maximum, local minimum, or neither. Answer: The critical points are x = -1 and x = 0. At x = -1, there is a local minimum, and at x = 0, there is a local maximum.

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

To find the first derivative of f(x), we'll need to use the product rule. Since f(x) = x^2 * sqrt(x+1), the product rule is: $$\frac{d}{dx}(u\cdot v) = u'v + uv'$$ where u = x^2 and v = sqrt(x+1). First, find the derivative of u and v: $$u' = \frac{d}{dx}(x^2) = 2x$$ $$v' = \frac{d}{dx}(\sqrt{x+1}) = \frac{1}{2\sqrt{x+1}}$$ Now, apply the product rule and simplify: $$f'(x) = u'v + uv' = 2x\sqrt{x+1} + x^2\frac{1}{2\sqrt{x+1}}$$

Step 2: Find the critical points.

To find the critical points, we'll need to solve the equation f'(x) = 0 or where f'(x) is undefined: $$2x\sqrt{x+1} + \frac{x^2}{2\sqrt{x+1}}=0$$ Factor out \(\sqrt{x+1}\): $$\sqrt{x+1}(2x + \frac{x^2}{2\sqrt{x+1}})=0$$ The critical points are when x = -1, and x = 0.

Step 3: Determine the domain.

The domain of f(x) is given in the problem as the closed interval [-1, 1]. Thus, we only need to evaluate the local behavior of the function within this domain.

Step 4: Use a graphing utility to determine local max, min, or neither.

Using a graphing utility, we can observe the behavior of the function f(x) at the critical points x = -1 and x = 0: - At x = -1, f(x) has a local minimum. - At x = 0, f(x) has a local maximum. In summary: Critical points x = -1, corresponds to a local minimum, and x = 0 corresponds to a local maximum on the interval [-1, 1].

Key Concepts

Product Rule

Calculating derivatives can be a bit tricky, especially when you have a function that's the product of two simpler functions. This is where the product rule comes in handy. The product rule is a fundamental tool in calculus. It is used to find the derivative of two differentiable functions multiplied together. If you have a function of the form \( f(x) = u(x) \cdot v(x) \), the product rule states that the derivative \( f'(x) \) is given by:

• \( f'(x) = u'(x)v(x) + u(x)v'(x) \)

This formula allows us to take the derivative of each function separately and then combine them.
In our exercise, \( u(x) = x^2 \) and \( v(x) = \sqrt{x+1} \). By finding \( u'(x) = 2x \) and \( v'(x) = \frac{1}{2\sqrt{x+1}} \), and applying the product rule, we effectively find the derivative of the entire function. This is crucial for identifying the behavior of the function, such as critical points.

Local Maximum

In calculus, a local maximum of a function is a point where the function value is higher than nearby points. It's like climbing to the top of a hill, but there might be higher peaks elsewhere.
Local maximum points are identified by examining the first derivative. If \( f'(x) \) changes from positive to negative at a critical point, that's typically where a local maximum occurs.

• Critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined.

For the function \( f(x) = x^2 \sqrt{x+1} \), we have a critical point at \( x = 0 \). Using a graphing utility, we confirm that the function changes from increasing to decreasing at this point, indicating a local maximum. This visual and analytical verification helps ensure our understanding is accurate.

Local Minimum

A local minimum is the opposite of a local maximum; it's a point where the function value is lower than nearby points, like reaching the bottom of a valley. To find a local minimum, we again rely on the derivative.

• If \( f'(x) \) changes from negative to positive, the function has a local minimum.

In the provided problem, at the critical point \( x = -1 \), the function \( f(x) \) dips lower than at nearby points. Using the graphing utility, this point is a local minimum because the graph shows the function transitioning from decreasing to increasing. This shift in behavior at the critical point is a clear indicator of a local minimum.

Graphing Utility

A graphing utility, whether a sophisticated calculator or computer software, is an invaluable tool for visualizing functions. It allows us to see the graph of a function, making it easier to identify features like local maxima and minima.

• By plotting \( f(x) = x^2 \sqrt{x+1} \) over the interval \([-1, 1]\), we can observe the critical behavior visually.

The graph provides a visual confirmation that at \( x = 0 \), the function reaches a peak, corroborating our analytical findings of a local maximum. Similarly, at \( x = -1 \), the graph indicates a trough, confirming the local minimum. Graphing utilities thus serve as a powerful method to complement calculus methods, offering clarity and confidence in analyzing functions.