Problem 7 - Active Calculus

Question

Let \(p\) be a function whose second derivative is \(p^{\prime \prime}(x)=(x+1)(x-2) e^{-x}\). a. Construct a second derivative sign chart for \(p\) and determine all inflection points of \(p\). b. Suppose you also know that \(x=\frac{\sqrt{5}-1}{2}\) is a critical number of \(p\). Does \(p\) have a local minimum, local maximum, or neither at \(x=\frac{\sqrt{5}-1}{2}\) ? Why? c. If the point \(\left(2, \frac{12}{e^{2}}\right)\) lies on the graph of \(y=p(x)\) and \(p^{\prime}(2)=-\frac{5}{e^{2}},\) find the equation of the tangent line to \(y=p(x)\) at the point where \(x=2\). Does the tangent line lie above the curve, below the curve, or neither at this value? Why?

Step-by-Step Solution

Answer

a) Inflection points at \(x = -1\) and \(x = 2\). b) Neither local minimum nor maximum at \(x = \frac{\sqrt{5}-1}{2}\). c) Tangent line: \(y = -\frac{5}{e^{2}}x + \frac{22}{e^{2}}\) lies below the curve at \(x = 2\).

Step 1: Finding Critical Points for Second Derivative

Determine where the second derivative changes sign. Set \(p''(x) = 0\) and solve: \[(x+1)(x-2)e^{-x} = 0\] Since \(e^{-x} > 0\) for all \(x\), we solve: \[(x+1)(x-2) = 0\] Then, the critical points are \(x = -1\) and \(x = 2\).

Step 2: Constructing the Second Derivative Sign Chart

Analyze the sign of \(p''(x)\) using the intervals \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\): For \(x e -1\) and \(x e 2\), \((x+1) > 0\) when \(x > -1\), \((x-2) > 0\) when \(x > 2\). Therefore, \(p''(x)\) changes sign at both \(x = -1\) and \(x = 2\).

Step 3: Identifying Inflection Points

Since \(p''(x)\) changes sign at \(x = -1\) and \(x = 2\), these points are inflection points.

Step 4: Analyzing Local Extrema

Given that \(x = \frac{\sqrt{5}-1}{2}\) is a critical number, determine the concavity: Use the second derivative sign chart: For \(x = \frac{\sqrt{5}-1}{2} \ \text{(which lies between -1 and 2)}\), \(p''(x)< 0\). Therefore, \(x = \frac{\sqrt{5}-1}{2}\) is neither local minimum nor maximum.

Step 5: Finding the Equation of the Tangent Line

Given \(p(2) = \frac{12}{e^{2}}\) and \(p'(2) = -\frac{5}{e^{2}}\), the equation of the tangent line at \(x = 2\) is: \[y - \frac{12}{e^{2}} = -\frac{5}{e^{2}}(x - 2)\] Simplify: \[y = -\frac{5}{e^{2}}x + \frac{22}{e^{2}}\]

Step 6: Position of the Tangent Line Relative to the Curve

Given that \(p'(2) = -\frac{5}{e^{2}} < 0\), the tangent line lies below the curve at \(x = 2\) since the function is decreasing at that point.

Key Concepts

Inflection Points

Inflection points are critical in understanding the behavior of a function. An inflection point is where the function changes its concavity, i.e., from concave up to concave down, or vice versa. To find inflection points, we use the second derivative of the function, noted as \( p''(x) \).
In our exercise, we are given \( p''(x) = (x + 1)(x - 2)e^{-x} \). To determine where \( p''(x) \) changes sign, we set the second derivative equal to zero and solve for \( x \):

• \((x + 1)(x - 2) = 0 \)
• The solutions are \( x = -1 \) and \( x = 2 \)

These points are where the second derivative can change its sign. We then analyze intervals around these points to construct a sign chart. Since the sign of \( p''(x) \) changes at both \( x = -1 \) and \( x = 2 \), we determine that these are inflection points.

Critical Points

Critical points of a function specify where the first derivative \( p'(x) \) is zero or undefined. These points can indicate potential local maxima, minima, or points of inflection. However, in part of our exercise, you know that \( x = \frac{\sqrt{5}-1}{2} \) is a critical number of \( p(x) \).
To determine if this point is a local extremum, we analyze the concavity at this critical point using the second derivative sign chart. Since \( x = \frac{\sqrt{5}-1}{2} \) lies between \( -1 \) and \( 2 \), and \( p''(x) < 0 \) in this interval, \( x = \frac{\sqrt{5}-1}{2} \) is classified as neither a local minimum nor a local maximum.

Tangent Line Equation

Finding the equation of the tangent line to a function at a given point is a fundamental concept in calculus. The slope of the tangent line at a point \( x = a \) on the function is given by \( p'(a) \). The equation of the tangent line can be written using the point-slope form:
\[ y - p(a) = p'(a) (x - a) \] In our exercise, we are given \( p(2) = \frac{12}{e^2} \) and \( p'(2) = -\frac{5}{e^2} \). Therefore, the equation of the tangent line at \( x = 2 \) is:
\[ y - \frac{12}{e^2} = -\frac{5}{e^2} (x - 2) \] Simplifying, we get: \[ y = -\frac{5}{e^2} x + \frac{22}{e^2} \] This line represents the tangent to the curve at \( x = 2 \). Given that the slope is negative, indicating that the tangent line lies below the curve at \( x = 2 \).

Local Extrema

Local extrema refer to local maxima and minima, points where the function takes on a highest or lowest value compared to its immediate surroundings. Typically, we use the first derivative \( p'(x) \) and critical points to determine these values.
In our problem, it is noted that \( x = \frac{\sqrt{5}-1}{2} \) is a critical point, but by examining the second derivative sign chart, we see that \( p''(x) < 0 \) in the interval \( -1 < x < 2 \). This negative value indicates a change in concavity, confirming that \( x = \frac{\sqrt{5}-1}{2} \) is neither a local maximum nor a local minimum. Therefore, it is simply a point of inflection.

Concavity

Concavity describes the manner in which a curve bends. A function is concave up where its second derivative \( p''(x) \) is positive, making the curve bend upwards. Conversely, it is concave down where \( p''(x) \) is negative, making the curve bend downwards.
To determine concavity, we analyze signs of the second derivative \( p''(x) \). As per our exercise, we have:\( p''(x) = (x + 1)(x - 2)e^{-x} \), and we study it over intervals \( (-\infty, -1) \), \( (-1, 2) \), and \( (2, \infty) \):

• For \( (-\infty, -1) \), both \( (x + 1) \) and \( (x - 2) \) are negative. Therefore, \( p''(x) > 0 \), and the function is concave up.
• For \( (-1, 2) \), \( (x + 1) > 0 \) and \( (x - 2) < 0 \). Thus, \( p''(x) < 0 \), and the function is concave down.
• For \( (2, \infty) \), both \( (x + 1) \) and \( (x - 2) \) are positive, yielding \( p''(x) > 0 \), making the function concave up.

This understanding helps in visualizing the function's graph and its inflection points better.