Problem 31 - 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: Find the derivative of the function

The given function is \(f(x) = x e^{a x}\). To find the critical points, we need to determine its derivative. Utilize the product rule for differentiation, \((uv)' = u'v + uv'\), where \(u = x\) and \(v = e^{a x}\). This gives us \(f'(x) = (1)e^{ax} + x(a e^{ax})\), simplifying to \(f'(x) = e^{ax} + a x e^{ax}\).

Step 2: Set the derivative equal to zero

Critical points occur where the derivative is zero or undefined. Set \(f'(x) = e^{ax} + a x e^{ax} = 0\). Factor out \(e^{ax}\) to get \(e^{ax}(1 + ax) = 0\). Since \(e^{ax}\) is never zero, we must have \(1 + ax = 0\).

Step 3: Solve for the parameter \(a\)

Given that the critical point is at \(x = 3\), substitute \(x = 3\) into the equation \(1 + ax = 0\). This yields \(1 + 3a = 0\). Solve for \(a\) to get \(3a = -1\), hence \(a = -\frac{1}{3}\).

Key Concepts

Derivative Calculation

When you are working with calculus problems involving critical points, the first essential step is to find the derivative of the given function.
To find the derivative, you apply differentiation rules, which help you determine how the function changes at each point. In simpler terms, you're finding a new function that tells you the slope (rate of change) of the original function at any point.
In this particular exercise, we were given a function, \(f(x) = x e^{a x}\). The derivative of this function, \(f'(x)\), is crucial because it helps identify where the function increases or decreases. \[ f'(x) = e^{ax} + ax e^{ax} \] This formula was derived using the product rule, which is a special method explained in the next section.

Product Rule

The product rule is a technique used in calculus for finding the derivative of a product of two functions.
When faced with a function composed of two differentiable functions multiplied together, like \(u(x)\) and \(v(x)\), you apply the rule:

• First, take the derivative of the first function \(u\), and multiply it by the second function \(v\).
• Then, take the original function \(u\) and multiply it by the derivative of the second function \(v\).

Add these two results together to create the derivative of the product. Mathematically expressed as \((uv)' = u'v + uv'\) .
In our problem, \(u = x\) and \(v = e^{ax}\). Thus, applying the product rule gave us: \[ f'(x) = (1)e^{ax} + x(a e^{ax}) = e^{ax} + ax e^{ax} \]Understanding the product rule is key for differentiating complex functions that involve multiplications of variable parts.

Critical Point Analysis

Finding critical points involves determining where the derivative of a function equals zero or becomes undefined. Critical points signify potential minimums or maximums in the function.
In our exercise, the function was \(f(x) = x e^{a x}\) and its derivative: \[ f'(x) = e^{ax} + ax e^{ax} \]This derivative needs to be set equal to zero to find the critical points.
Setting the equation \(e^{ax}(1 + ax) = 0\) helps identify values of \(x\) where critical points occur, provided \(e^{ax}\) never equals zero. Therefore, we consider \(1 + ax = 0\).
To find the specific factor \(a\) that makes \(x=3\) a critical point, solve \(1 + 3a = 0\). The solution \(a = -\frac{1}{3}\) confirms that substituting \(a\) into the original equation gives us a point where the slope changes its behavior, marking it as a critical point.