Problem 60 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the inflection point(s), if any, of each function. $$ f(x)=x e^{-2 x} $$

Step-by-Step Solution

Answer

The inflection point of the function \(f(x) = xe^{-2x}\) occurs at \(\left(\frac{3}{2}, \frac{3}{2}e^{-3}\right)\).

Step 1: Find the first derivative f'(x)

To find the first derivative of the function f(x) = \(xe^{-2x}\), we'll apply the Product Rule for differentiation (u(x)v(x))' = u'(x)v(x) + u(x)v'(x). In this case, u(x) = x and v(x) = \(e^{-2x}\). First, find the derivative of u(x): \[\frac{d}{dx}(x) = 1\] Next, find the derivative of v(x): \[\frac{d}{dx}(e^{-2x}) = -2e^{-2x}\] Now apply the Product Rule: \[f'(x) = (1)(e^{-2x}) + (x)(-2e^{-2x})\] This will simplify to: \[f'(x) = e^{-2x}(1 - 2x)\]

Step 2: Find the second derivative f''(x)

Now we will find the second derivative by differentiating f'(x) with respect to x. Since f'(x) is also a product of two functions, we need to apply the Product Rule again. Let u(x) = \(e^{-2x}\) and v(x) = \(1 - 2x\). Differentiate u(x): \[\frac{d}{dx}(e^{-2x}) = -2e^{-2x}\] Differentiate v(x): \[\frac{d}{dx}(1 - 2x) = -2\] Now apply the Product Rule: \[f''(x) = (-2e^{-2x})(1 - 2x) + (e^{-2x})(-2)\]

Step 3: Set the second derivative f''(x) equal to 0 and solve for x

To find the points of inflection, we need to solve the equation f''(x) = 0. Let's simplify f''(x) first: \[f''(x) = -2e^{-2x} (1 - 2x) - 2e^{-2x} = -2e^{-2x} (3 - 2x)\] Now set f''(x) equal to 0: \[-2e^{-2x}(3 - 2x) = 0\] Divide both sides by -2: \[e^{-2x}(3 - 2x) = 0\] This equation is true if either of these factors equals zero: Factor 1: \(e^{-2x} = 0\) --This is never true for any value of x. The exponential function is always non-zero. Factor 2: \(3 - 2x = 0\)

Step 4: Solve for x

Now, solve the equation \(3 - 2x = 0\) for x: \[3 - 2x = 0\] \[2x = 3\] \[x = \frac{3}{2}\] So, there is one inflection point at \(x = \frac{3}{2}\).

Step 5: Find the value of the function at the inflection point

To find the value of the function at the inflection point, substitute the value of x back into the original function f(x) = \(xe^{-2x}\): \[f\left(\frac{3}{2}\right) = \frac{3}{2}e^{-2\cdot\frac{3}{2}} = \frac{3}{2}e^{-3}\] Hence, the function experiences an inflection point at \(\left(\frac{3}{2}, \frac{3}{2}e^{-3}\right)\).

Key Concepts

Inflection Points

Inflection points are a fascinating concept in calculus where the curvature of a function changes direction. They tell us where a function goes from concave up (curving upwards) to concave down (curving downwards), or vice versa.
This makes them essential for understanding the shape of a graph.

To find inflection points, we look at the second derivative of the function. The second derivative, usually denoted as \(f''(x)\), provides information on how the slope changes.
If \(f''(x)\) changes sign from positive to negative or negative to positive, there is an inflection point at that \(x\) value.

Here's a quick guide on finding them:

• First, find the first derivative \(f'(x)\). This gives us a function representing the gradient or slope of \(f(x)\).
• Then, differentiate again to obtain the second derivative \(f''(x)\).
• Set \(f''(x) = 0\) and solve for \(x\).
• Confirm an inflection point by checking if \(f''(x)\) indeed changes sign at the solutions.

This methodology helps in locating the inflection points, enhancing our understanding of the function's overall behavior on its domain.

Differential Calculus

Differential calculus is a central pillar of calculus that focuses on the concept of the derivative.
It allows us to analyze the behavior of functions in terms of rates of change and slopes of curves.

The derivative is a fundamental concept in differential calculus. It measures how a function value changes as its input changes slightly.
This is analogous to determining the speed of a car at an instant, instead of its average speed over time.
In mathematical terms, it's the limit of the average rate of change as the interval approaches zero.

Here are some key steps in differentiating a function:

• Identify the function you want to differentiate.
• Apply differentiation rules such as the power rule, product rule, quotient rule, or chain rule depending on the function's structure.
• Simplify the obtained derivative to better understand the behavior of the function.

In the context of our exercise, the Product Rule is particularly important as it helps in finding the derivative of products of functions.
Understanding and applying these rules unveils important information about the function's slopes and concavity.

Product Rule

The Product Rule is a special technique in differential calculus for deriving the derivative of a product of two functions.
Understanding this rule is crucial when you face functions that are multiplied together.

Mathematically, if you have two functions \(u(x)\) and \(v(x)\), then the derivative of their product \(u(x)v(x)\) is given by:
\[(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\]
This rule takes into account both functions separately, maintaining the accuracy of the rate of change.

Here's how it works:

• First, find the derivative of the first function \(u(x)\).
• Keep the second function \(v(x)\) as it is and multiply by the derivative of the first function.
• Then, find the derivative of the second function \(v(x)\).
• Keep the first function \(u(x)\) as it is and multiply by the derivative of the second function.
• Add these two results to obtain the derivative of the product.

For example, in the exercise, the function \(f(x) = xe^{-2x}\) required the product rule because \(x\) and \(e^{-2x}\) are multiplied together.
The strategic use of the Product Rule allows us to break down complex product functions into simpler parts for differentiation.