Problem 44 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

\(y=\left(x^{2}-x\right) e^{-x}\) where \(x=0\)

Step-by-Step Solution

Answer

The value of the function at \( x = 0 \) is 0.

Step 1: Understand the Function

The given function is \( y = (x^2 - x) e^{-x} \). We need to evaluate this function at \( x = 0 \).

Step 2: Substitute \( x = 0 \) into the Function

Substitute \( x = 0 \) into the function: \( y = (0^2 - 0) e^{-0} \).

Step 3: Simplify the Expression

Simplify the expression: \( y = (0 - 0) e^{0} = 0 \times 1 = 0 \).

Key Concepts

substitution in calculus

Substitution in calculus is a fundamental technique used for evaluating functions at specific points.
Let's break it down using the provided exercise.

In the function given, \( y = (x^2 - x) e^{-x} \), we need to evaluate it at \( x = 0 \).

This means we're substituting \( x = 0 \) into the function. Substitution is simple: wherever you see \( x \), replace it with the value provided.

So, substituting \(x=0\) results in: \( y = (0^2 - 0) e^{-0} \). This simplifies the function to a specific value at \( x=0 \).

Now that the substitution is done, it makes further operations easier, turning the complex function into something manageable.

simplifying expressions

Simplifying expressions is a crucial step in solving math problems. It involves reducing the expression to its simplest form.

After substituting \(x=0\) in the function \( y = (x^2 - x) e^{-x} \), we get \( y = (0^2 - 0) e^{-0} \).

First, let's simplify inside the parenthesis: \( 0^2 = 0 \) and \( 0 - 0 = 0 \).
So, now the function looks like: \( y = 0 \times e^{-0} \).

Next, we simplify the exponential part: \( e^{-0} \). Any number to the power of 0 is 1, hence \( e^{0} = 1 \).

Then, multiply the simplified results: \( y = 0 \times 1 = 0 \).

This step-by-step simplification is what makes the final evaluation straightforward.

exponential functions

Exponential functions are a type of function where a constant base is raised to a variable exponent.

In the given function, \( y = (x^2 - x) e^{-x} \), \( e^{-x} \) is the exponential part.

Exponential functions are notable because of their unique properties:

• They grow (or decay) at rates proportional to their current value.
• \( e \) is a mathematical constant approximately equal to 2.718.

An important property of exponentials is that \( e^0 = 1 \).

In our example, this simplifies calculations significantly when \( x = 0 \). It turns \( e^{-0} = e^{0} = 1 \).

This highlights the power of exponentials in transforming the complexity of functions.