Problem 22 - Applied Calculus

Question

(a) Use a graph to estimate the \(x\) -values of any critical points and inflection points of \(f(x)=e^{-x^{2}}\) (b) Use derivatives to find the \(x\) -values of any critical points and inflection points exactly.

Step-by-Step Solution

Answer

Critical point at \( x = 0 \); Inflection points at \( x = \pm \frac{1}{\sqrt{2}} \).

Step 1: Identify the Function

We are given the function \( f(x) = e^{-x^2} \). Our goal is to identify critical points and inflection points.

Step 2: Graphical Estimation of Critical Points

To estimate the critical points, plot the function \( f(x) = e^{-x^2} \). Look for the points where the slope of the graph is zero, which are typically the peaks or valleys of the graph. This is often where the `turning points` are visible.

Step 3: Graphical Estimation of Inflection Points

From the graph of \( f(x) = e^{-x^2} \), observe where the concavity changes. These changes occur where the curves shift from concave up to concave down or vice versa. Estimate these \( x \)-values.

Step 4: Find First Derivative for Critical Points

Differentiate the function: \( f'(x) = \frac{d}{dx}[e^{-x^2}] = -2xe^{-x^2} \). Critical points occur where this derivative is zero or undefined. Set \( f'(x) = 0 \): \(-2xe^{-x^2} = 0\). Solve for \( x \) to find that the critical point is at \( x = 0 \).

Step 5: Find the Second Derivative for Inflection Points

Differentiate the first derivative to find the second derivative: \( f''(x) = \frac{d}{dx}[-2xe^{-x^2}] = 2e^{-x^2} - 4x^2e^{-x^2} \). Simplify this to \( f''(x) = e^{-x^2}(2 - 4x^2) \).

Step 6: Solve for Inflection Points Using the Second Derivative

Inflection points occur where \( f''(x) = 0 \). Set \( e^{-x^2}(2 - 4x^2) = 0 \). Since \( e^{-x^2} eq 0 \) for any real \( x \), solve for \( 2 - 4x^2 = 0 \). This gives \( x^2 = \frac{1}{2} \), so \( x = \pm \frac{1}{\sqrt{2}} \).

Key Concepts

Inflection Points

Inflection points are places on a curve where the concavity changes. Concavity refers to how the curve bends, either opening upwards or downwards. Think about it like this: when a curve changes from looking like a "cup" (concave up) to a "cap" (concave down), you have crossed an inflection point.
To accurately determine inflection points, we use the second derivative of a function. Specifically, inflection points occur where the second derivative changes sign. For our function, we calculated the second derivative and found it to be:

• \(f''(x) = e^{-x^2}(2 - 4x^2)\)

To find the inflection points, solve the equation \(2 - 4x^2 = 0\) for \(x\). Solving gives \(x^2 = \frac{1}{2}\), so the x-values are \(x = \pm \frac{1}{\sqrt{2}}\). These are the inflection points, clearly showing where the concavity changes.

Derivative Calculation

Derivative calculation is crucial for understanding changes in the shape and direction of graphs. Derivatives tell us about the slope or steepness of a function at any point. The first derivative, in particular, lets us find critical points, which are where slopes are zero or undefined. Careful calculation is needed at this step.
The function we worked on is \( f(x) = e^{-x^2} \). To find the critical points, we first calculated its first derivative, which came out to be:

• \(f'(x) = -2xe^{-x^2}\)

Critical points are found by setting this derivative equal to zero and solving for \(x\). In this case:

• \(-2xe^{-x^2} = 0\)

This simplifies to \(x = 0\), indicating the critical point is at \(x = 0\). This calculation shows the importance of derivatives in identifying key features of functions.

Graphical Estimation

Graphical estimation helps in visualizing mathematical concepts and is a useful first step when analyzing functions. Before diving into detailed calculations, it’s often beneficial to get a visual sense by plotting the function. With our function \(f(x) = e^{-x^2}\), graphing illustrates where the slope levels off, hinting at critical points, and where the concavity changes, suggesting inflection points.
By observing the graph:

• Critical points become apparent at the peaks or dips, which can initially be estimated, as in our example at \(x = 0\).
• Inflection points appear where the bending of the graph changes direction, initially observed around \(x = \pm \frac{1}{\sqrt{2}}\).

These graphical insights offer a practical way of narrowing down where to focus detailed calculations, enhancing understanding beyond algebraic methods.