Problem 44 - Precalculus: Functions and Graphs

Question

Solve the system for \(a\) and \(b\). (Hint: Treat terms such as \(e^{3 x}, \cos x,\) and \(\sin x\) as "constant coefficients.") $$\left\\{\begin{aligned} a e^{-x}+b e^{4 x} &=0 \\ -a e^{-x}+b\left(4 e^{4 x}\right) &=2 \end{aligned}\right.$$

Step-by-Step Solution

Answer

\(a = -\frac{2}{5} e^x, b = \frac{2}{5 e^{4x}}\).

Step 1: Identify and Eliminate Terms

First, we recognize each equation in the system. For each, terms involving different expressions of the variable should be isolated: \( a e^{-x} + b e^{4x} = 0 \) and \(-a e^{-x} + 4 b e^{4x} = 2 \). To eliminate terms, we add the two equations to eliminate \(a e^{-x}\).

Step 2: Simplify the System

Add the two equations: \((a e^{-x} + b e^{4x}) + (-a e^{-x} + 4b e^{4x}) = 0 + 2 \). This reduces to \(5b e^{4x} = 2\).

Step 3: Solve for Coefficient b

Divide both sides of the equation by \(5 e^{4x}\) to isolate \(b\): \(b = \frac{2}{5 e^{4x}}\).

Step 4: Substitute to Find Coefficient a

With \(b = \frac{2}{5 e^{4x}}\), substitute back into the first equation: \(a e^{-x} + \frac{2}{5 e^{4x}} e^{4x} = 0\). Simplify to get \(a e^{-x} + \frac{2}{5} = 0\).

Step 5: Solve for Coefficient a

Solve for \(a\): \(a e^{-x} = -\frac{2}{5}\). Dividing by \(e^{-x}\) gives \(a = -\frac{2}{5} e^x\).

Key Concepts

Exponential Functions

An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In the exercise provided, functions like \( e^{-x} \) and \( e^{4x} \) are exponential functions because they're expressed as powers of the base \( e \).
This fascinating function type plays a pivotal role in many scientific fields, modeling exponential growth or decay.

• The notation \( e \) refers to Euler's number, approximately equal to 2.71828, which is a mathematical constant.
• The variability of the exponent, such as \( -x \) or \( 4x \), dictates whether the function grows or decays. Negative exponents, like \( e^{-x} \), indicate decay, whereas positive exponents, like \( e^{4x} \), indicate growth.

Understanding these functions involves recognizing their general behavior and how they differ from more linear or polynomial functions. This knowledge becomes crucial when solving equations that incorporate exponential elements.In solving systems with exponential functions, keep in mind that we treat them as if they were coefficients when they multiply unknowns like \( a \) and \( b \). This perspective simplifies the process and allows us to use standard algebraic procedures.

Solving Equations

Solving equations involves finding the value(s) of variables that make an equation true.
In the reviewed exercise, this process is applied to a system of equations, wherein two separate equations are solved at once. Here's how:

• We start by understanding each equation: \(a e^{-x} + b e^{4x} = 0\) and \(-a e^{-x} + 4b e^{4x} = 2\).
• The next step is elimination. By adding both equations, one can eliminate terms strategically — in this case, \( a e^{-x} \).
• After performing the elimination, we reduce the system to a manageable form — here, leading to \(5b e^{4x} = 2\).

Once we find \(b\), we can substitute its value back into one of the initial equations to find \(a\). The substitution method effectively helps isolate variables one at a time, simplifying the solving process even in complex scenarios, such as with exponential terms.

Coefficient Manipulation

Coefficient manipulation is the process of transforming and rearranging equations to simplify and solve them. It is particularly important in systems with varying coefficients, needing a keen understanding of how to handle each term.
In the original exercise, terms like \(e^{-x}\) and \(e^{4x}\), although not constants in a traditional sense, function as coefficients for \(a\) and \(b\). This quirky setup might seem confusing, but treating these terms as constants allows straightforward manipulation.

• First, isolate terms to streamline the equation, such as adding equations to eliminate \(a e^{-x}\) in our problem.
• Once simplified, real coefficients (\(b\)) can be isolated by division.
• Post-isolation, substitute the found coefficients (values) back into the initial setup to find the remaining unknowns (in this case, \(a\)).

The idea is to carefully align equations to make them simpler through strategic addition, subtraction, and factor division, leading to the coefficient solutions. By understanding these techniques, solving complex systems becomes far less daunting.