Problem 10 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=x /(x-4) $$

Step-by-Step Solution

Answer

The function is concave up on \((4, \infty)\) and concave down on \((-\infty, 4)\). There are no points of inflection or critical points due to the asymptote at \(x=4\).

Step 1: Find the First Derivative

The first step is to find the derivative of the function \( f(x) = \frac{x}{x-4} \) using the quotient rule. The quotient rule states that \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = x \) and \( v = x-4 \). So, \( u' = 1 \) and \( v' = 1 \). Applying the quotient rule, we get:\[f'(x) = \frac{(1)(x-4) - (x)(1)}{(x-4)^2} = \frac{x-4-x}{(x-4)^2} = \frac{-4}{(x-4)^2}\]

Step 2: Find the Second Derivative

Next, we calculate the second derivative of the function to determine concavity. Start by differentiating \( f'(x) = \frac{-4}{(x-4)^2} \). Use the power rule and chain rule:\[f''(x) = \frac{d}{dx} \left[ -4(x-4)^{-2} \right] = 8(x-4)^{-3} = \frac{8}{(x-4)^3}\]

Step 3: Determine Concavity and Points of Inflection

The concavity of a function is determined by the sign of its second derivative:- The function is **concave up** where \( f''(x) > 0 \).- It is **concave down** where \( f''(x) < 0 \).- Points of inflection occur where \( f''(x) = 0 \) or where \( f''(x) \) changes sign. Examine the expression \( \frac{8}{(x-4)^3} \):- If \( x > 4 \), \( (x-4)^3 > 0 \) so \( f''(x) > 0 \), implying concave up.- If \( x < 4 \), \( (x-4)^3 < 0 \) so \( f''(x) < 0 \), implying concave down.- There's a vertical asymptote at \( x = 4 \), so no point of inflection.

Step 4: Find Critical Points Using First Derivative

A critical point occurs where \( f'(x) = 0 \) or is undefined. From \( f'(x) = \frac{-4}{(x-4)^2} \), we see that \( f'(x) \) is never zero. However, \( f'(x) \) is undefined at \( x = 4 \), reinforcing the vertical asymptote.

Step 5: Apply the Second Derivative Test

We use the Second Derivative Test to classify critical points as local minimums or maximums. Since \( f'(x) \) is not zero anywhere and \( f''(x) = \frac{8}{(x-4)^3} \) changes sign around \( x = 4 \), there's no local min or max at \( x = 4 \) since it's not a defined point on \( f \). Thus, no local min/max exists on this typical interval without further specific constraints on domain.

Key Concepts

First Derivative

Let's start with the first derivative, a powerful tool for understanding the behavior and properties of a function. The first derivative of a function, denoted as \( f'(x) \), provides valuable insights into the slope at any given point on the function's graph.

To find the first derivative of the function \( f(x) = \frac{x}{x-4} \), we employ the quotient rule. The quotient rule is a technique used to differentiate functions that are divided by each other. According to this rule, if you have a function \( \frac{u}{v} \), its derivative is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Here, \( u = x \) and \( v = x-4 \). This results in:\[ f'(x) = \frac{(1)(x-4) - (x)(1)}{(x-4)^2} = \frac{-4}{(x-4)^2} \]

The first derivative tells us about the increasing or decreasing nature of the function. Since \( f'(x) = \frac{-4}{(x-4)^2} \) is always negative, this reflects that the function is always decreasing, except where undefined.

Second Derivative

After finding the first derivative, the next step is to calculate the second derivative. The second derivative, \( f''(x) \), provides information about the concavity of a function. It shows us how the rate of change of the function’s slope itself changes.

For \( f(x) = \frac{x}{x-4} \), the second derivative is derived by differentiating \( f'(x) = \frac{-4}{(x-4)^2} \). Here we use the power and chain rules:

• Using power rule: Treat \( -4(x-4)^{-2} \) as \( -4 imes (x-4)^{-2} \)
• Using chain rule: Derivative of \( (x-4)^{-2} \) is \( -2(x-4)^{-3} \)

Thus, we get:\[ f''(x) = 8(x-4)^{-3} = \frac{8}{(x-4)^3} \]

The sign of \( f''(x) \) determines the concavity:

• If \( f''(x) > 0 \), the graph is concave up.
• If \( f''(x) < 0 \), the graph is concave down.

Critical Points

Critical points are vital in analyzing a function's behavior where potential local maxima or minima might occur. These points are found by identifying where the first derivative \( f'(x) \) is equal to zero or where the derivative does not exist. They indicate points where the function may change direction from increasing to decreasing, or vice versa.

With our function \( f'(x) = \frac{-4}{(x-4)^2} \), notice that it never actually equals zero because the numerator, \(-4\), is a constant. However, the derivative is undefined at \( x = 4 \), where the denominator becomes zero. This implies a vertical asymptote at \( x = 4 \).

As a result, \( x = 4 \) is a critical point in terms of the function's undefined behavior rather than a potential root for a maxima or minima. Since it's a vertical asymptote and not a typical y-coordinate point on the graph, there are no changes in direction here.

Second Derivative Test

The Second Derivative Test is a useful method to classify critical points further as locations of local maxima or minima. Once critical points are identified, the second derivative helps determine their nature.

For the function \( f(x) = \frac{x}{x-4} \), since our first derivative is non-zero at defined points, and the second derivative \( f''(x) = \frac{8}{(x-4)^3} \) has a sign change around the point \( x = 4 \), it suggests transitions in concavity at the neighborhood about this undefined point.

• If \( f''(x) > 0 \) surrounding a critical point, it implies a local minimum.
• If \( f''(x) < 0 \) surrounding a critical point, it implies a local maximum.

However, since \( x = 4 \) isn't actually in the domain of the function, we don't mark any local maxima or minima here. The undefined nature at \( x = 4 \) prevents a solid conclusion on local extremal behavior from this test alone in this context.