Problem 75 - Precalculus with Limits

Question

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. \(-x e^{-x}+e^{-x}=0\)

Step-by-Step Solution

Answer

The solution to the equation \(-x e^{-x}+e^{-x}=0\) is \(x=1\)

Step 1 - Simplify the equation

Looking closely at the equation, we can see it is possible to factor out \(e^{-x}\) from both terms of the equation. This will simplify the equation to: \(e^{-x}(-x+1)=0\)

Step 2 - Solve for the variable

In this step, set \(-x+1=0\) to solve for \(x\). Add \(x\) to both sides and subtract 1 from both sides to get: \(x=1\).

Step 3 - Rounding

Round off the solution of \(x\) to three decimal places. However, as the solution is \(1\), which doesn't have any decimal places, there's no need to do any rounding in this case. Therefore, \(x\) remains as \(1\).

Step 4 - Verification

Verify the answer using a graphing utility. Plot the function \(-x e^{-x}+e^{-x}\) on the graphing utility. The point where the curve intersects the x-axis represents the solution to the equation which should be \(x=1\).

Key Concepts

Factoring Exponential Functions

Factoring is a powerful technique that simplifies complex mathematical expressions. In the given problem, the equation \[-x e^{-x}+e^{-x}=0\]involves the exponential function \(e^{-x}\), which can be factored out because it appears in both terms. This transforms the expression into:\[e^{-x}(-x+1)=0\].
Factorization involves recognizing common terms across elements of an equation, allowing you to rewrite these equations in a simpler, equivalent form. This step is essential as it can often reveal the straightforward path to finding solutions.

• Identify common factors: Both terms here contain \(e^{-x}\), showing it can be simplified further.
• Rewrite the equation: By factoring, we express it as \(e^{-x}(-x+1)=0\).
• Solve each part: Since \(e^{-x}\) is never zero, set \(-x+1=0\) to find the solution for \(x\).

This type of rewriting through factoring is particularly useful, enabling easier manipulation and solution of otherwise complicated exponential equations.

Graphical Verification

Visualizing mathematical problems can be tremendously helpful when solving equations, especially when verifying solutions. Using a graphing utility to verify the solution of the equation \[-x e^{-x}+e^{-x}=0\]can provide you with a clear understanding. Plotting the function, you'll notice the points where it crosses the x-axis, corresponding to the equation’s solutions.
Graphical verification steps include:

• Plot the function: Input the function into a graphing tool to visualize it over a chosen range of x-values.
• Observe the intersection: Look for the point where the curve crosses the x-axis. This intercept point corresponds to the value of \(x\) that satisfies the equation.
• Compare with algebraic solution: Ensure the graphical solution at \(x=1\) matches your algebraic calculation.

Graphing not only reassures your solution but also helps build an intuition about the behavior of mathematical functions. It aids in understanding the relationship between components visually.

Simplifying Equations

Simplifying equations is the cultivation of reduced complexity from initial conditions to theoretical solutions. In the given equation \[-x e^{-x}+e^{-x}=0\],the first step involves reducing it to a simpler form by factoring. When simplified to \[e^{-x}(-x+1)=0\],the equation becomes more manageable. Solving simpler equations precisely gives clearer paths to potential solutions.
Key points in simplifying include:

• Factor common elements: Identify terms that can be factored, such as \(e^{-x}\) in this case.
• Solve the reduced component: After simplifying, focus on solving the condensed component of the equation \(-x+1=0\).
• Reduce operations: Focus on minimal arithmetic steps necessary, which helps in maintaining accuracy and efficiency.

Simplification helps in eliminating unnecessary complexity, narrowing down to logical decision points. It enables more straightforward arithmetic and algebraic manipulations, leading to faster solution procurement.