Question

A function \(f(x)\) is given. (a) Find the possible points of inflection of \(f\). (b) Create a number line to determine the intervals on which \(f\) is concave up or concave down. \(f(x)=x^{2} \ln x\)

Short answer

The possible inflection point is at \(x = e^{-\frac{3}{2}}\). \(f(x)\) is concave down on \((0, e^{-\frac{3}{2}})\) and concave up on \((e^{-\frac{3}{2}}, +\infty)\).

Step-by-step solution

Identify the second derivative

First, we need to find the second derivative of the function to determine concavity and possible inflection points. The function is \(f(x) = x^2 \ln x\). Start by finding the first derivative, \(f'(x) = (x^2)' \ln x + x^2 (\ln x)'\). This simplifies to \(f'(x) = 2x \ln x + x\). Next, find the second derivative by differentiating \(f'(x)\):\[f''(x) = (2x \ln x)' + (x)' = 2 \ln x + 2 + 1\].Thus, \(f''(x) = 2 \ln x + 3\).

Find points where the second derivative is zero or undefined

Points of inflection occur where the second derivative is zero or undefined. Solve \(f''(x) = 0\):\[2 \ln x + 3 = 0\],\[2 \ln x = -3\],\[\ln x = -\frac{3}{2}\],\[x = e^{-\frac{3}{2}}\].The function \(f(x)\) is undefined for \(x \leq 0\) because \(\ln x \) is undefined for non-positive \(x\). Thus, \(x = e^{-\frac{3}{2}}\) is a possible inflection point.

Create a number line

Create a number line to analyze the concavity intervals around the possible inflection point at \(x = e^{-\frac{3}{2}}\). Test intervals around \(x = e^{-\frac{3}{2}}\), such as \((0, e^{-\frac{3}{2}})\) and \((e^{-\frac{3}{2}}, +\infty)\).

Test concavity with a number line

Choose a test point in each interval and substitute it into the second derivative \(f''(x) = 2 \ln x + 3\):* For interval \((0, e^{-\frac{3}{2}})\), choose \(x = 0.11\). Here, \(f''(0.11) = 2 \ln(0.11) + 3 \) which is negative, indicating \(f(x)\) is concave down.* For interval \((e^{-\frac{3}{2}}, +\infty)\), choose \(x = 1\). Here, \(f''(1) = 2 \ln(1) + 3 = 3 \), which is positive, indicating \(f(x)\) is concave up.

Key concepts

Second Derivative

The second derivative of a function, often denoted as \( f''(x) \), gives us insight into the function's concavity, which describes the direction that a curve bends. To find the second derivative, we start by determining the first derivative. For the given function \( f(x) = x^2 \ln x \), the first derivative \( f'(x) \) is found using the product rule of calculus. The product rule helps differentiate functions that are multiplied together.After finding \( f'(x) = 2x \ln x + x \), the next step is to differentiate \( f'(x) \) again to get the second derivative. Applying differentiation rules to \( f'(x) \) results in \( f''(x) = 2 \ln x + 3 \). The second derivative tells us how the rate of change of the first derivative is itself changing. In other words, it provides information about the "sharpness" of the curve.

Concavity

Concavity describes whether the graph of a function is bending upwards or downwards. The second derivative plays a critical role in determining concavity. If \( f''(x) > 0 \) at a point, the function is concave up (like a cup). If \( f''(x) < 0 \), the function is concave down (like a frown).In our exercise, we find the second derivative \( f''(x) = 2 \ln x + 3 \). By testing values within specific intervals, we can determine where the function is concave up or down. Testing points within the interval \((0, e^{-\frac{3}{2}})\), such as \( x = 0.11 \), we find \( f''(0.11) \) is negative, indicating the function is concave down in this interval. For \( x > e^{-\frac{3}{2}} \), the concavity changes as \( f''(1) \) is positive, indicating concave up.

First Derivative

The first derivative of a function, \( f'(x) \), represents the rate of change or the slope of the function at any point. It's the first step in understanding how the function behaves.To find the first derivative of \( f(x) = x^2 \ln x \), we use the product rule. This rule is needed because the function is a product of \( x^2 \) and \( \ln x \). According to the product rule, the derivative of \( u \cdot v \) is \( u' \cdot v + u \cdot v' \), where \( u = x^2 \) and \( v = \ln x \). Calculating these, we find that \( f'(x) = 2x \ln x + x \). The first derivative tells us the direction—whether the function is increasing or decreasing—which aids in determining extrema points and preparing to find points of inflection.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is a mathematical function that helps us analyze the behavior of functions involving exponential growth or decay. It’s defined only for positive values of \( x \) because \( \ln x \) is undefined for zero and negative values.In this exercise, the natural logarithm is combined with a power of \( x \), resulting in the function \( f(x) = x^2 \ln x \). When we differentiate \( \ln x \), using the rule \( (\ln x)' = \frac{1}{x} \), its presence significantly influences the derivatives and the overall shape of the graph of \( f(x) \). For instance, when solving \( 2 \ln x + 3 = 0 \) for the inflection point, knowing the properties of the natural logarithm allows us to find \( x \) efficiently, leading to the solution \( x = e^{-\frac{3}{2}} \). Understanding how \( \ln x \) affects differentiation is crucial for finding changes in concavity and behavior of the function.