Problem 29 - 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. $$ f(x)=x^{2}-8 \ln (x) $$

Step-by-Step Solution

Answer

The function is concave up for all \(x > 0\); no concave down intervals or inflection points. Local minimum at \(x = 2\).

Step 1: Find the First and Second Derivatives

To analyze concavity and critical points, we first find the derivative, \( f'(x) \), to determine critical points. Then, calculate \( f''(x) \) to assess concavity and use the second derivative test.For \( f(x) = x^2 - 8 \ln(x) \):- The first derivative: \( f'(x) = 2x - \frac{8}{x} \).- The second derivative: \( f''(x) = 2 + \frac{8}{x^2} \).

Step 2: Determine Critical Points

To find the critical points, set \( f'(x) = 0 \):\[ 2x - \frac{8}{x} = 0 \].Solving for \(x\):Multiply through by \(x\) to clear the fraction:\[ 2x^2 = 8 \]\[ x^2 = 4 \]\[ x = \pm 2 \].However, since \(x\) must be positive due to the \( \ln(x) \) term, only \(x = 2\) is considered.

Step 3: Test for Concavity Using the Second Derivative

Use \( f''(x) \) to determine intervals of concavity:- \( f''(x) = 2 + \frac{8}{x^2} \) is always positive for \( x > 0 \) since both terms are positive.Thus, \( f(x) \) is concave up for all \(x > 0\). There are no intervals where the function is concave down.

Step 4: Determine Points of Inflection

Points of inflection occur where \( f''(x) = 0 \) or changes sign, indicating changes in concavity. Since \( f''(x) = 2 + \frac{8}{x^2} \) is always positive in its domain, there are no points where it changes sign.Thus, there are no points of inflection.

Step 5: Apply the Second Derivative Test for Local Extrema

The Second Derivative Test uses \( f''(x) \) at the critical point \( x = 2 \):- \( f''(2) = 2 + \frac{8}{4} = 4 \), which is positive.This indicates that \( f(x) \) has a local minimum at \( x = 2 \).

Key Concepts

Concavity

When talking about concavity, we are referring to the direction that a curve opens. A function is said to be "concave up" if it curves upwards like a cup or a smiley face. We often visualize this concept as a bowl or a valley. Conversely, a function is "concave down" when it curves downwards, resembling a frown or an upside-down bowl.
In the context of calculus, concavity is determined using the second derivative \( f''(x) \). A positive second derivative (\( f''(x) > 0 \)) indicates that the function is concave up over that interval. Similarly, a negative second derivative (\( f''(x) < 0 \)) shows that the function is concave down.
For the function \( f(x) = x^2 - 8 \ln(x) \), when the second derivative \( f''(x) = 2 + \frac{8}{x^2} \) is assessed, it is always positive for all \( x > 0 \). Therefore, the function is always concave up for all positive \( x \).

• **Key points:**
• **Concave up:** Function opens upwards, \( f''(x) > 0 \)
• **Concave down:** Function opens downwards, \( f''(x) < 0 \)

Critical Points

Critical points are where the function's derivative is zero or undefined. These points are crucial because they usually indicate potential maxima, minima, or points of inflection on the function graph.
To find critical points, we set the first derivative \( f'(x) \) equal to zero. For our function \( f(x) = x^2 - 8 \ln(x) \), we previously found that \( f'(x) = 2x - \frac{8}{x} \). Setting this equal to zero provides the equation \( 2x = \frac{8}{x} \).
Solving this, we multiply both sides by \( x \), resulting in \( 2x^2 = 8 \), which simplifies to \( x = \pm 2 \). However, due to the presence of the \( \ln(x) \) term, only positive \( x \) values are valid, thus leaving \( x = 2 \) as the only critical point for the function in its domain.

Points of Inflection

Points of inflection are where the graph of a function changes its concavity from up to down or down to up. These points are identified by where the second derivative is zero or changes sign.
For our function, however, the second derivative \( f''(x) = 2 + \frac{8}{x^2} \) is positive for all \( x > 0 \). This means that \( f(x) \) never changes from concave up to concave down, or vice versa, and hence, there are no points of inflection.
In essence, for an inflection point to exist, the sign of \( f''(x) \) must change as \( x \) goes through a point, which doesn’t happen in our case.

• **Key points:**
• **Inflection point occurs** when \( f''(x) \) changes sign.
• No inflection points:** If \( f''(x) \) does not change sign.

Local Extrema

Local extrema refer to the local maximum and minimum values of a function. These are points where the function reaches a peak (maximum) or a trough (minimum).
To identify local extrema, we can use the Second Derivative Test at the critical points. For \( f(x) = x^2 - 8 \ln(x) \), we've already determined there is a critical point at \( x = 2 \).
By evaluating the second derivative \( f''(x) \) at \( x = 2 \), we find \( f''(2) = 2 + \frac{8}{4} = 4 \). Since this result is positive, the function has a local minimum at \( x = 2 \).

• **Key points:**
• **Local maximum:** If \( f''(x) < 0 \) at the critical point.
• **Local minimum:** If \( f''(x) > 0 \) at the critical point, as is the case here.