Question

Graphing functions Make a complete graph of the following functions using the graphing guidelines outlined. $$p(x)=x e^{-x^{2} / 2}$$

Short answer

Answer: The function has a local maximum at $$x=0$$ and local minima at $$x=\pm1$$.

Step-by-step solution

Determine the Domain and Range of the function

The Domain refers to all the possible values of $$x$$ for which the function is defined. Since the function is continuous for all real numbers (no restrictions on $$x$$), its domain is: $$(-\infty, \infty)$$ The Range refers to all possible values of $$p(x)$$. Since $$e^{-x^2 / 2}$$ is always positive and the maximum value of $$e^{-x^2 / 2}$$ is 1 (when $$x=0$$), the range will be from $$(0,\infty)$$.

Find X and Y intercepts

To find the X intercept, we need to find the points at which the graph of the function intersects the X-axis. In other words, when $$p(x)=0$$. $$p(x)=xe^{-x^2 / 2} = 0$$ The function is zero only when $$x=0$$. X-intercept = 0 To find the Y intercept, we need to find the points at which the graph of the function intersects the Y-axis (i.e. when $$x=0$$). $$p(0)=0\cdot e^{-0} = 0$$ Y-intercept = 0

Find Local Maxima and Minima

To find the local maxima and minima, we need to find the critical points of the function. Critical points occur when the derivative $$p'(x)$$ is zero or undefined. First, compute the derivative of $$p(x)$$ with respect to $$x$$ $$p'(x) = (xe^{-x^2/2})'$$ Using the product rule, we get: $$p'(x) = e^{-x^2 /2} + x(-x)e^{-x^2 /2}$$ Now, to find the critical points, set $$p'(x)$$ to zero: $$e^{-x^2 /2} - x^2 e^{-x^2 /2} = 0$$ Factor out the term $$e^{-x^2/2}$$: $$e^{-x^2 /2}(1 - x^2)=0$$ This equation is zero only when $$x=0$$ or $$x=\pm 1$$. After further analyzing, we will notice that at $$x = 0$$, $$p'(x)$$ is a maximum while at $$x = \pm 1$$, it is a minimum.

Identify Asymptotes and Behavior at End Points

There are no asymptotes for this function, as there are no points where the function approaches infinity or approaches a particular value without actually reaching it. To find the behavior at end points, observe the function as $$x$$ approaches $$-\infty$$ and $$+\infty$$: - As $$x \rightarrow -\infty$$, the term $$e^{-x^2/2}$$ will approach 1 while the term $$x$$ will approach $$-\infty$$. Therefore, the function approaches $$-\infty$$. - As $$x \rightarrow +\infty$$, the term $$e^{-x^2/2}$$ will approach 0 while the term $$x$$ will approach $$+\infty$$. Therefore, the function approaches 0.

Draw the Graph

Now, with all the previously computed information, draw the graph of the function: 1. Plot the X and Y-intercepts (0,0). 2. Plot the maximum point (0,0) and the minimum points ($$\pm1, \pm \frac{e^{-1/2}}{2}$$). 3. From the analysis of the behavior at the end-points, draw the function approaching $$-\infty$$ as $$x$$ goes to $$-\infty$$ and getting very close to 0 as $$x$$ goes to $$+\infty$$. 4. Join all the points together, forming a smooth curve. After drawing the graph, you will notice that the function has a local maximum at $$x=0$$, local minima at $$x=\pm1$$, and the graph is symmetric around the Y-axis.

Key concepts

Understanding Domain and Range

When it comes to graphs, the domain and the range are essential concepts you need to master for a clear visualization of functions. The **domain** of a function signifies all the possible input values (x-values) for which the function is defined. For the function \( p(x) = x e^{-x^{2}/2} \), there are no restrictions on \( x \), meaning you can plug in any real number into the function. Therefore, the domain is: \((-\infty, \infty)\).
On the other hand, the **range** represents all the possible output values (y-values) the function can produce. Since the term \(e^{-x^2 / 2}\) is always positive, the output of \(p(x)\) ranges from just above 0 to positive infinity. Hence, the range for this function is: \((0, \infty)\).
Understanding domain and range helps you set the stage for correctly sketching and analyzing graphs.

Identifying Intercepts

Intercepts are key points where the graph crosses the axes, providing useful reference points. The **X-intercept** is found where the graph crosses the X-axis, meaning the function equals zero at this point. For \( p(x) = x e^{-x^2 / 2} \), setting it equal to zero shows the function only equals zero at \( x = 0 \). Thus, the X-intercept is (0,0).
Similarly, the **Y-intercept** is where the graph crosses the Y-axis. This happens when \( x = 0 \). Substituting \( x = 0 \) into \( p(x) \), we also find the Y-intercept as (0,0). Both intercepts are the same point, which offers a good starting point for drawing the graph.

• X-intercepts occur where the function is zero, so solve \( p(x) = 0 \).
• Y-intercepts are found by evaluating \( p(x) \) at \( x = 0 \).

Intercepts form a basis for understanding the graph's roots and crossings.

Exploring Critical Points

A function's critical points, where its derivative is zero or undefined, reveal potential peaks and valleys in the graph—known as local maxima and minima. Here's a brief walkthrough on finding them for \( p(x) = x e^{-x^{2}/2} \).
First, derive the function: \(p'(x) = e^{-x^2/2} (1 - x^2)\).
Next, set \(p'(x) = 0\) to find critical points: \( e^{-x^2/2} (1 - x^2) = 0\). Since \( e^{-x^2/2} \) never reaches zero, focus on \( 1 - x^2 = 0 \). Solving it gives critical points \( x = 0 \) and \( x = \pm 1 \).
Further analysis shows:

• At \( x = 0 \), it's a local maximum.
• At \( x = \pm 1 \), they are local minima.

Critical points tell us where the graph changes direction and define the shape of the curve.

Clarifying Asymptotes

For many functions, asymptotes indicate where the function approaches a specific value but doesn’t actually reach it, often as it heads towards positive or negative infinity. For our function \( p(x) = x e^{-x^{2}/2} \), there are no vertical or horizontal asymptotes, as there are no limitations or divergences at infinity.
The behavior at infinity, however, still provides insight into its shape:

• As \( x \rightarrow -\infty \), \( p(x) \rightarrow -\infty \).
• As \( x \rightarrow +\infty \), \( p(x) \rightarrow 0 \), showing it flattens out near the X-axis.

Knowing that the function approaches but never reaches certain limits helps in sketching the end behavior of the graph, ensuring more accurate representation of its behavior.