Problem 34 - Applied Calculus

Question

Find the value of \(a\) so that the function \(f(x)=x e^{a x}\) has a critical point at \(x=3\)

Step-by-Step Solution

Answer

The value of \(a\) is \(-\frac{1}{3}\).

Step 1: Identify the condition for critical points

A critical point occurs where the derivative of the function is zero or undefined. In this case, you want to find when the derivative of \(f(x) = x e^{ax}\) is zero, given that the critical point is at \(x = 3\).

Step 2: Compute the derivative of the function

The function is \(f(x) = x e^{ax}\). Use the product rule, \((uv)' = u'v + uv'\), where \(u = x\) and \(v = e^{ax}\). Thus, the derivatives are \(u' = 1\) and \(v' = ae^{ax}\). Therefore, \(f'(x) = 1 \cdot e^{ax} + x \cdot ae^{ax} = e^{ax} + axe^{ax}\).

Step 3: Set the derivative equal to zero

Set \(f'(x) = e^{ax} + axe^{ax} = 0\) since the critical point is at \(x = 3\).

Step 4: Simplify the equation for the critical point

Factor out \(e^{ax}\) from the equation: \(e^{ax}(1 + ax) = 0\). Since \(e^{ax}\) is never zero, the equation simplifies to \(1 + ax = 0\).

Step 5: Solve for \(a\) at \(x = 3\)

Substitute \(x = 3\) into the equation \(1 + ax = 0\), giving \(1 + a \cdot 3 = 0\). Solve for \(a\) to get \(a = -\frac{1}{3}\).

Key Concepts

calculus problems

Calculus problems often involve finding and analyzing critical points. These critical points can reveal important characteristics about the behavior and properties of a function. In particular, critical points occur where a function's derivative is zero or undefined. This might sound a bit tricky at first, but it just means we're looking for the points where the slope of the function flattens out.

Finding critical points is crucial because they help identify local maximums, minimums, or points of inflection—all useful for understanding how a function behaves. This task typically involves three main steps:

• Find the derivative of the given function.
• Set the derivative equal to zero to locate potential critical points.
• Solve these equations for the variable, often denoted as \(x\).

With these processes, calculus gives us a toolkit to delve deeply into the behavior of various functions.

product rule

The product rule is an essential tool in calculus when dealing with functions that are multiplied together. Suppose you have a function that is the product of two other functions, denoted as \(u\) and \(v\). The derivative of this product is not simply the product of the derivatives of \(u\) and \(v\). Instead, you use the product rule, which states that:\[(uv)' = u'v + uv'\]Here's how it works:

• The first step is to differentiate the first function \(u\) while keeping the second function \(v\) unchanged.
• The second step involves keeping the first function \(u\) unchanged and differentiating the second function \(v\).
• Finally, add the results from these differentiations to get the derivative of the product.

In the case of the function \(f(x) = x e^{ax}\), you assign \(u = x\) and \(v = e^{ax}\). Then, compute the derivatives as \(u' = 1\) and \(v' = ae^{ax}\), followed by applying the product rule to find \(f'(x) = e^{ax} + axe^{ax}\). This step is key in derivative problems where two functions are combined.

derivative analysis

Derivative analysis is a powerful method used to understand how a function behaves. Once you have the derivative of a function, you analyze it to determine where a function increases or decreases and identify its critical points. Let's explore how this study is done effectively:1. **Take the Derivative:** Begin with the derivative of a function. For our example, the derivative was found as \(f'(x) = e^{ax} + ax e^{ax}\). This expression combines terms where both the original function and its derivative parts appear.
2. **Set the Derivative to Zero:** You set \(f'(x) = 0\) to find the \(x\)-values where the slope of the function is zero. In mathematical problems like this, these intersections offer points where the function changes direction from increasing to decreasing or vice versa.
3. **Solve and Understand:** By solving the equation \(e^{ax}(1 + ax) = 0\), trivial solutions arise. Since the exponential function \(e^{ax}\) is never zero, we only need to solve \(1 + ax = 0\). This simplifies further calculation and narrows down the critical point effectively. Through derivative analysis, we can deduce and solve for specific values, such as \(a\), which ensures the critical point's position as described in the original setup.