Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the critical points and use the test of your choice to decide which critical points give a local maximum value and which give a local minimum value. What are these local maximum and minimum values? $$ f(x)=\frac{x^{2}}{\sqrt{x^{2}+4}} $$

Step-by-Step Solution

Answer

The critical point is at \( x = 0 \), which is a local minimum with a value of 0.

Step 1: Find the Derivative

To find the critical points, we first need to find the derivative of the function. The function is given by \[ f(x) = \frac{x^2}{\sqrt{x^2+4}}. \] We will use the quotient rule, where if \( u = x^2 \) and \( v = \sqrt{x^2 + 4} \), then the derivative \( f'(x) \) is given by \[ f'(x) = \frac{u'v - uv'}{v^2}. \] Calculate each part: \[ u' = 2x, \quad v = (x^2 + 4)^{1/2}, \quad v' = \frac{x}{\sqrt{x^2+4}}. \] Substitute these into the formula: \[ f'(x) = \frac{2x(x^2+4)^{1/2} - x^2 \frac{x}{\sqrt{x^2+4}}}{x^2+4}. \] Simplify to obtain: \[ f'(x) = \frac{x(2(x^2+4) - x^2)}{(x^2+4)^{3/2}}. \] Further simplify: \[ f'(x) = \frac{x(8)}{(x^2+4)^{3/2}} = \frac{8x}{(x^2 + 4)^{3/2}}. \]

Step 2: Find the Critical Points

Critical points occur where the derivative \( f'(x) = 0 \) or is undefined. The derivative \( f'(x) = \frac{8x}{(x^2 + 4)^{3/2}} \) is defined for all real \( x \) as the denominator is never zero (since \( x^2 + 4 \geq 4 \)). Set \[ \frac{8x}{(x^2 + 4)^{3/2}} = 0, \] to find zeroes, solve \[ 8x = 0 \Rightarrow x = 0. \] Thus, the only critical point is \( x = 0 \).

Step 3: Use the Second Derivative Test

To determine the nature of the critical point, compute the second derivative \( f''(x) \). First, differentiate \( f'(x) = \frac{8x}{(x^2+4)^{3/2}} \):Use the product rule and chain rule: \[ u = 8x, \quad u' = 8, \quad v = (x^2+4)^{-3/2}, \quad v' = -\frac{3x}{(x^2+4)^{5/2}}. \]Then, \[ f''(x) = 8(x^2+4)^{-3/2} + \frac{8x \cdot (-3x)}{(x^2+4)^{5/2}}. \]Simplify to find \[ f''(x) = \frac{8(x^2+4) - 24x^2}{(x^2+4)^{5/2}}. \]Evaluating at the critical point \( x = 0 \):\[ f''(0) = \frac{32}{4^{5/2}} = \frac{32}{8} = 4 > 0. \]Since \( f''(0) > 0 \), \( x = 0 \) is a local minimum.

Step 4: Calculate the Local Minimum Value

The local minimum occurs at \( x = 0 \). Substitute back into the original function to find the minimum value:\[ f(0) = \frac{0^2}{\sqrt{0^2+4}} = 0. \]Therefore, the local minimum value of the function is 0. There is no local maximum since the derivative did not indicate any other critical points.

Key Concepts

Derivative

In calculus, the derivative of a function represents the rate at which the function’s value changes with respect to a change in its input value. It is a fundamental concept used to find critical points, which are key to turning points in a function. For the function \( f(x) = \frac{x^2}{\sqrt{x^2+4}} \), the first step in analyzing it is to find its derivative. This tells us where the function is increasing or decreasing.

To find the derivative, we apply the quotient rule, which is used when differentiating a function that is the quotient of two other functions. This rule states that if \( u(x) \) and \( v(x) \) are functions, the derivative \( f'(x) \) of \( \frac{u}{v} \) is:

• \( f'(x) = \frac{u'v - uv'}{v^2} \)

By substituting \( u = x^2 \) and \( v = \sqrt{x^2+4} \), we compute the derivative:

• \( u' = 2x \)
• \( v' = \frac{x}{\sqrt{x^2+4}} \)
• \( f'(x) = \frac{8x}{(x^2 + 4)^{3/2}} \)

This expression for the derivative is crucial in finding the critical points of the function.

Local Maximum

A local maximum occurs at a point in a function if the function value at that point is greater than at points immediately surrounding it. For a critical point to be a local maximum, the function must change from increasing to decreasing. In mathematical terms, this is confirmed if the second derivative at that point is negative.

In this particular exercise, the derivative \( f'(x) = \frac{8x}{(x^2 + 4)^{3/2}} \) does not lead to any points where \( f'(x) \) changes from positive to negative. This indicates that there are no local maxima in this function, as no such critical points exist where the derivative behavior matches that of a local maximum.

Since the second derivative test or any changes in the first derivative do not support the existence of a local maximum in this instance, the function \( f(x) \) solely presents a local minimum.

Local Minimum

Local minima in a function are points where the value of the function is lower than at any nearby points. To find and confirm a local minimum, we check the critical points found through the derivative calculation and use the second derivative test.

For \( f(x) = \frac{x^2}{\sqrt{x^2+4}} \), the sole critical point occurs at \( x = 0 \). The second derivative \( f''(x) \) is evaluated at this point, and if \( f''(0) > 0 \), it confirms a local minimum.

• \( f''(0) = 4 > 0 \).

This positive value confirms that \( x = 0 \) is indeed a local minimum. Additionally, substituting back \( x = 0 \) into the original function \( f(0) = 0 \) gives us the local minimum value, which is 0.

Quotient Rule

The quotient rule is a method used in differentiation when a function is expressed as a fraction of two other functions. It allows us to find the derivative of the quotient \( \frac{u}{v} \) given the derivatives of \( u \) and \( v \). This is especially useful for functions like \( f(x) = \frac{x^2}{\sqrt{x^2 + 4}} \), where both the numerator and the denominator are functions of \( x \).

To apply the quotient rule, remember the formula:

• \( \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \)

In this context, applying the quotient rule helped us determine \( f'(x) \), which we then used to locate critical points. The correct application of the quotient rule is essential in ensuring the right results for complex expressions, supporting the overall analysis of the function's behavior.