Problem 38 - Applied Calculus

Question

Find constants \(a\) and \(b\) in the function \(f(x)=a x e^{b x}\) such that \(f\left(\frac{1}{3}\right)=1\) and the function has a local maximum at \(x=\frac{1}{3}\)

Step-by-Step Solution

Answer

\(a = 3e, b = -3\)

Step 1: Understand the function

We are given the function \(f(x) = a x e^{b x}\) and need to find the constants \(a\) and \(b\) so that it satisfies the conditions: 1) \( f\left(\frac{1}{3}\right) = 1\) and 2) the function has a local maximum at \(x = \frac{1}{3}\).

Step 2: Apply given function value condition

Substitute \(x = \frac{1}{3}\) into the function: \(f\left(\frac{1}{3}\right) = a \cdot \frac{1}{3} \cdot e^{b \cdot \frac{1}{3}} = 1\). Solve for \(a\) in terms of \(b\): \(a \cdot \frac{1}{3} \cdot e^{\frac{b}{3}} = 1\), leading to \(a = \frac{3}{e^{\frac{b}{3}}}\).

Step 3: Find the derivative to locate critical points

The derivative of \(f(x) = a x e^{b x}\) is found using the product and chain rules: \(f'(x) = a e^{b x} + a x b e^{b x}\). Factor out common terms: \(f'(x) = a e^{b x}(1 + bx)\).

Step 4: Set the derivative to zero at local maximum

Set \(f'(x) = 0\) to find critical points: \(a e^{b x}(1 + bx) = 0\). Since \(a e^{b x} eq 0\), we have \(1 + bx = 0\). Substituting \(x = \frac{1}{3}\) implies \(1 + \frac{b}{3} = 0\) leading to \(b = -3\).

Step 5: Determine constant \(a\)

Substitute \(b = -3\) back into the equation for \(a\): \(a = \frac{3}{e^{\frac{-3}{3}}} = 3 e\).

Step 6: Verify the solution

Substitute \(a = 3e\) and \(b = -3\) back into the original function equation and check the conditions: 1. \(f\left(\frac{1}{3}\right) = 3e \cdot \frac{1}{3} \cdot e^{-1} = 1\), which holds true.2. Check second derivative \(f''(x)\) to confirm local maximum at \(x = \frac{1}{3}\). We find \(f''\left(\frac{1}{3}\right) < 0\), confirming a local maximum.

Key Concepts

Local Maximum

Understanding "local maximum" is an essential part of solving calculus problems. A local maximum is a point on a function's graph where the value of the function is greater than all other nearby points.
In simple terms, think of it as the "peak" in its immediate neighborhood.
To determine if a function has a local maximum at a certain point, we typically utilize the derivative of the function. If the derivative changes sign from positive to negative at a point, the function has a local maximum at that point.

• This indicates a rise to the high point and then a fall afterward.

However, it is also crucial to evaluate the second derivative. The condition for a local maximum is reinforced if the second derivative at the point is negative.
That means, the curve is concave down, verifying that it is at the "top" of a hill.

Derivative

The derivative is a fundamental concept in calculus. It provides the rate at which a function is changing at any given point. For example, if you think of the function as a moving car, the derivative at any point tells you the speed of the car right then.

• It helps understand how a function behaves and changes over its domain.
• This change in behavior can help locate critical points like local maximums or minimums.

In this context, finding where the derivative equals zero helps locate these critical points. These are the places where the function's slope is flat—either maximum, minimum, or a point of inflection. To differentiate our given function, which involves a product of functions, we apply the product and chain rules of differentiation to find the derivative.

Calculus Problem Solving

Solving calculus problems involves a combination of techniques and understanding multiple concepts. First, understanding the problem requirements: what values are needed, and what conditions the function should meet. In our example, we are asked to find the constants for the function such that it has a specific function value at a point and a local maximum at the same point. This involves:

• Substituting known values into the function to create equations.
• Using derivatives to find critical points where changes in behavior occur.
• Applying the second derivative test to confirm the nature of these critical points.

Step-by-step approaches, like the one used to solve this particular problem, help in breaking down complex problems into manageable parts.
It involves systematically applying calculus principles to derive the solution.