Problem 26 - 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. $$ f(x)=x \mathrm{e}^{-x} $$

Step-by-Step Solution

Answer

Concave down on \((-\infty, 2)\); concave up on \((2, \infty)\); inflection point at \(x=2\); local maximum at \(x=1\).

Step 1: Find the First Derivative

Start by finding the first derivative of the function \( f(x) = x e^{-x} \). Use the product rule for differentiation: \((u\cdot v)' = u'v + uv'\), where \(u = x\) and \(v = e^{-x}\). The derivatives are \(u' = 1\) and \(v' = -e^{-x}\). Hence:\[f'(x) = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x} - xe^{-x} = e^{-x}(1 - x).\]

Step 2: Find the Second Derivative

Find the second derivative by differentiating \(f'(x) = e^{-x}(1 - x)\). Use the product rule again: \((u\cdot v)' = u'v + uv'\), where \(u = e^{-x}\) and \(v = 1 - x\). The derivatives are \(u' = -e^{-x}\) and \(v' = -1\). Thus:\[f''(x) = -e^{-x}(1 - x) + e^{-x}(-1) = e^{-x}(x - 2).\]

Step 3: Determine Concavity Intervals

Set the second derivative equal to zero to find points of inflection:\[e^{-x}(x - 2) = 0.\]The exponential term \(e^{-x}\) is never zero, so we solve \(x - 2 = 0\), giving \(x = 2\). Test intervals around \(x = 2\) to identify concavity:- For \(x < 2\), choose \(x = 1\): \(f''(1) = e^{-1}(1 - 2) < 0\); concave down.- For \(x > 2\), choose \(x = 3\): \(f''(3) = e^{-3}(3 - 2) > 0\); concave up.So, \(f(x)\) is concave down on \((-f, 2)\) and concave up on \((2,
6f)\). The point \(x = 2\) is a point of inflection.

Step 4: Find Critical Points

Find the critical points by setting the first derivative equal to zero:\[e^{-x}(1 - x) = 0.\]Since \(e^{-x} eq 0\), solve \(1 - x = 0\), obtaining \(x = 1\). This is the only critical point.

Step 5: Use the Second Derivative Test

Apply the Second Derivative Test at the critical point \(x = 1\). Calculate \(f''(1)\):\[f''(1) = e^{-1}(1 - 2) = -e^{-1} < 0.\]Since \(f''(1) < 0\), the function \(f(x)\) has a local maximum at \(x = 1\).

Key Concepts

Concave Up

A function is said to be concave up on an interval if its second derivative is positive for that interval. This means that the curve of the function is opening upwards, resembling a smile. In the context of our exercise, we determined

• the second derivative function to be \(f''(x) = e^{-x}(x - 2)\).
• To find where the function is concave up, we test intervals around the point \(x = 2\).
• For \(x > 2\), say \(x = 3\): \(f''(3) = e^{-3}(3 - 2) > 0\).

Hence, the function \(f(x) = x e^{-x}\) is concave up on the interval \((2, \infty)\). This indicates that for any \(x\) within this interval, the graph of the function has a curvature that opens upwards.

Concave Down

Conversely, a function is concave down on an interval if its second derivative is negative in that interval. This makes the curve look like a frown, curving downwards. By evaluating the second derivative of our function:

• We find that for \(x < 2\), like when \(x = 1\), \(f''(1) = e^{-1}(1 - 2) < 0\).

This indicates that the function \(f(x) = x e^{-x}\) is concave down on the interval \((-\infty, 2)\). The negative value of the second derivative here depicts that any point within this range presents the function's graph as having a downward opening curve.

Critical Points

Critical points of a function occur where its first derivative is equal to zero or is undefined. These points are crucial since they can indicate potential local maxima or minima. For our function:

• The first derivative is \(f'(x) = e^{-x}(1-x)\).
• Setting this equal to zero gives us \(1 - x = 0\), which leads to a critical point at \(x = 1\).

This point is significant in understanding where the function's behavior changes, indicating a potential peak or trough in the function's graph.

Points of Inflection

Points of inflection are where a function changes concavity, meaning it switches from concave up to concave down or vice versa. To find these points, we examine where the second derivative changes sign:

• The second derivative we calculated was \(f''(x) = e^{-x}(x - 2)\).
• To find inflection points, we solve \(e^{-x}(x - 2) = 0\), resulting in \(x = 2\).

At this point, our earlier analysis showed the function transitions from concave down to concave up. Therefore, \(x = 2\) is a point of inflection for the function \(f(x) = x e^{-x}\), marking a change in curvature on the graph.