Problem 32 - Single Variable Calculus

Question

In Exercises \(5-38\), sketch the graph of the function using the curve- sketching guidelines on page \(348 .\) $$ f(x)=x e^{x} $$

Step-by-Step Solution

Answer

The graph of the function \(f(x) = xe^x\) has the following properties: 1. Intercepts: It intersects the x-axis and y-axis at the point (0, 0). 2. Asymptotes: There is a horizontal asymptote at y = 0 but no vertical asymptotes. 3. Critical Point: The function has a critical point at x = -1, and it is decreasing for \(x < -1\) and increasing for \(x > -1\). 4. Inflection Point: There is an inflection point at x = -2, where the function changes from being concave down for \(x < -2\) to concave up for \(x > -2\). By considering this information, the graph can be sketched accordingly.

Step 1: 1. Intercepts

To find the x and y intercepts, we will solve the following equations: - For the x-intercept: \(f(x) = 0\). - For the y-intercept: \(f(0)\). X-intercept: \[ 0 = xe^x \] Since the only value for x that makes \(xe^x\) equal to zero is x = 0, we have one x-intercept: (0, 0). Y-intercept: \[ f(0) = 0 \times e^0 = 0 \] The y-intercept is the same point as the x-intercept: (0, 0).

Step 2: 2. Asymptotes

Vertical asymptotes occur when the function becomes infinitely large for some value of x. Since our function is a product of x and \(e^x\), there is no value of x for which the function becomes infinite. Thus, there are no vertical asymptotes. To check for horizontal asymptotes, we examine the limit as x approaches infinity and negative infinity: \[ \lim_{x\to\infty} xe^x = \infty \quad\text{and}\quad \lim_{x\to-\infty} xe^x = 0 \] As x approaches negative infinity, the function approaches 0, so there is a horizontal asymptote at y = 0.

Step 3: 3. First Derivative and Critical Points

To find the critical points, we will find the first derivative of f(x) and solve the equation: \(f'(x) = 0\). \[ f'(x) = \frac{d}{dx}(xe^x) = e^x + xe^x = e^x(1 + x) \] Now, let's set f'(x) to zero and solve: \[ e^x(1 + x) = 0 \] Since \(e^x\) is never zero, we can divide both sides by \(e^x\) and solve for x: \[ 1 + x = 0 \] \[ x = -1 \] The critical point is x = -1. Now, let's find the intervals of increase and decrease by analyzing the sign of f'(x) in the intervals \((-\infty, -1)\) and \((-1, \infty)\): - For \(x \in (-\infty, -1)\): \(f'(-2) = e^{-2}(1 - 2) = e^{-2}(-1) < 0\), so the function is decreasing in this interval. - For \(x \in (-1, \infty)\): \(f'(0) = e^0(1 + 0) = e^0 = 1 > 0\), so the function is increasing in this interval.

Step 4: 4. Second Derivative and Inflection Points

To find the points of inflection, we will find the second derivative of f(x) and solve the equation: \(f''(x) = 0\). \[ f''(x) = \frac{d^2}{dx^2}(xe^x) = \frac{d}{dx}(e^x + xe^x) = e^x(x + 2) \] Now, let's set f''(x) to 0 and solve: \[ e^x(x + 2) = 0 \] Since \(e^x\) is never zero, we can divide both sides by \(e^x\) and solve for x: \[ x + 2 = 0 \] \[ x = -2 \] The inflection point is x = -2. Now, let's find the intervals of concavity by analyzing the sign of f''(x) in the intervals \((-\infty, -2)\) and \((-2, \infty)\): - For \(x \in (-\infty, -2)\): \(f''(-3) = e^{-3}(-3 + 2) = -e^{-3} < 0\), so the function is concave down in this interval. - For \(x \in (-2, \infty)\): \(f''(0) = e^0(0 + 2) = 2e^0 = 2 > 0\), so the function is concave up in this interval. Now, we can use all the information we gathered to sketch the graph of the function \(f(x) = xe^x\). The graph will have a horizontal asymptote at y = 0, increase after the critical point at x = -1, and change concavity at the inflection point x = -2.

Key Concepts

Intercepts

Intercepts are essential features in curve sketching, helping us pinpoint where a graph crosses the coordinate axes. An intercept doesn't just give us a point; it provides a crucial guide for our sketch.

**X-Intercepts**: To find x-intercepts, set the function equal to zero. For the function \(f(x) = xe^x\), we solve \(xe^x = 0\). Here, canceling \(e^x\) (since it's never zero) leads to \(x = 0\). This means the graph crosses the x-axis at (0, 0).

**Y-Intercepts**: The y-intercept is found by inserting 0 into the function, giving \(f(0) = 0 \times e^0 = 0\). The graph crosses the y-axis at this same point (0, 0). This overlap of intercepts hints at a significant point for the graph's balancing act.

Asymptotes

Asymptotes help us visualize a function's behavior far out on the axes, often indicating the direction or height at extremes.

**Vertical Asymptotes**: These suggest where a function shoots to infinity. For \(f(x) = xe^x\), vertical asymptotes don't exist because no specific \(x\) value leads to an undefined function.

**Horizontal Asymptotes**: These show the function's behavior as \(x\) approaches infinity. For \(f(x) = xe^x\), observing the limit, \(\lim_{x\to\infty} xe^x = \infty\) and \(\lim_{x\to-\infty} xe^x = 0\). The latter indicates a horizontal asymptote at \(y = 0\), guiding us on how the function behaves towards negative infinity.

Critical Points

Critical points are the soul of curve sketching and help find where a function changes direction.

To discover these, we look at the first derivative. For \(f(x) = xe^x\), compute \(f'(x) = e^x(1 + x)\). Set this to zero: \(e^x(1 + x) = 0\). Simplifying, since \(e^x\) is never zero, we find \(1 + x = 0\), giving \(x = -1\).

At this critical point, the function is either at a peak or a valley. Testing intervals around \(-1\): areas where \(f'(x)\) is positive, the function rises; where negative, it falls. Thus, it decreases before \(-1\) and increases after it, confirming a minimum at \(x = -1\).

Inflection Points

Inflection points provide essential clues about a function's curvature, signaling concavity changes.

We locate inflection points by inspecting the second derivative. For our function, \(f''(x) = e^x(x + 2)\). Set \(f''(x) = 0\) gives \(x + 2 = 0\), leading to \(x = -2\).

This point signals a shift in the curvature. Testing intervals around \(-2\) shows us where the function bends. For \(x < -2\), \(f''(x) < 0\) implies concave down; for \(x > -2\), \(f''(x) > 0\) implies concave up. This change at \(x = -2\) gives a pivotal insight into the overall graph shape.