Problem 50 - Calculus an Applied Approach

Question

Use a symbolic differentiation utility to find the derivative of the function. Graph the function and its derivative in the same viewing window. Describe the behavior of the function when the derivative is zero. $$ f(x)=\sqrt{x}\left(2-x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x)=\sqrt{x}(2-x^{2})\) is \(-2x\sqrt{x}-\frac{(2-x^{2})}{2\sqrt{x}}\). The derivative equals to zero at \(x=0, \frac{2}{\sqrt{3}}\), which are the function's turning points.

Step 1: Differentiate The Function

Utilize the product rule and the chain rule for differentiation, which says that the derivative of a function of the form \(u(x)v(x)\) is \(u'(x)v(x) + u(x)v'(x)\). Differentiating the given function \(f(x)=\sqrt{x}(2-x^{2})\), results to: \(f'(x)=-2x\sqrt{x}-\frac{(2-x^{2})}{2\sqrt{x}}\).

Step 2: Find where the derivative is zero

Set the derivative equation to zero and solve for \(x\). This gives you the points on the \('x'\) axis where the slope of the function is zero. Solving \(-2x\sqrt{x}-\frac{(2-x^{2})}{2\sqrt{x}} = 0\), results in \(x=0, \frac{2}{\sqrt{3}}\).

Step 3: Graph the function and its derivative

Plot the original function \(f(x)=\sqrt{x}(2-x^{2})\) and its derivative \(-2x\sqrt{x}-\frac{(2-x^{2})}{2\sqrt{x}}\) on the same set of axis to visualize the behaviors of the function and its derivative.

Step 4: Analyze the Behavior of The Function

At \(x=0, \frac{2}{\sqrt{3}}\) where the derivative equals zero, the function reaches its maximum or minimum. Notice that at these points the function either increases then decreases (maximum point) or decreases then increases (minimum point).

Key Concepts

Product Rule

When dealing with calculus problems involving derivatives, one essential tool is the product rule. It's used specifically when differentiating functions that are products of two separate functions. Imagine you have two functions, say \( u(x) \) and \( v(x) \). The product rule lets you find the derivative of their product, \( u(x) v(x) \), by following this formula:

• First, differentiate \( u(x) \) to get \( u'(x) \).
• Next, differentiate \( v(x) \) to get \( v'(x) \).
• The derivative of the product is then \( u'(x) v(x) + u(x) v'(x) \).

In our exercise, we have \( f(x) = \sqrt{x}(2-x^2) \), where \( \sqrt{x} \) is \( u(x) \) and \( (2-x^2) \) is \( v(x) \). Apply the product rule with these components to find the derivative, which helps in understanding how the two parts of the function interact.

Chain Rule

Another critical concept in differentiation is the chain rule. This rule is particularly handy when dealing with composite functions, where one function is inside another. Think of it like peeling an onion, layer by layer.Say you have a function \( g(x) = h(f(x)) \). To find the derivative \( g'(x) \), you would:

• First, differentiate the outer function \( h(x) \) as if \( f(x) \) were just a simple variable. This gives you \( h'(f(x)) \).
• Then, multiply by the derivative of the inner function, \( f'(x) \).

In our problem, \( \sqrt{x} \) and \( 2-x^2 \) require careful handling using the chain rule, especially when combined with the product rule. The interplay of these two rules allows us to differentiate complex expressions involving nested functions.

Graphing Functions

Graphing is a visual tool that makes understanding functions and their derivatives much easier. To graph the function \( f(x) = \sqrt{x}(2-x^2) \) along with its derivative, it's crucial to plot the graphs on the same set of axes. This provides a clear view of how the function behaves as \( x \) changes.Steps to graph effectively:

• Identify the function components that might affect its shape, such as maxima, minima, and points of inflection.
• Calculate and mark these critical points where the derivative equals zero.
• Plot the function, noting changes in the graph which are reflected by corresponding critical points in the derivative.
• Superimpose the derivative to see where the slope of the function is increasing, decreasing, or constant.

By observing the plotted function and its derivative, you gain insight into the relationship between the function's shape and changes in its derivative.

Critical Points

Critical points are the values of \( x \) where the derivative of a function, \( f'(x) \), is zero or undefined. These points can indicate where a function reaches its highest or lowest values, known as maxima or minima.How to find and analyze critical points:

• First, take the derivative of the function to find \( f'(x) \).
• Set \( f'(x) = 0 \) and solve for \( x \). These solutions give you potential critical points.
• Study the function's behavior around these points. If the function switches from increasing to decreasing, it’s a local maximum. If it switches from decreasing to increasing, it’s a local minimum.
• In some cases, a critical point might be a saddle point, where the function does not have a local maximum or minimum but the tangent is horizontal.

For our function, solving \( -2x\sqrt{x} - \frac{(2-x^2)}{2\sqrt{x}} = 0 \) leads to critical points at \( x=0 \) and \( x=\frac{2}{\sqrt{3}} \). Analyzing these points helps understand how the function behaves at those values and why those points are significant in the graph.