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Chapter 4: Problem 44

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

Short Answer

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

Step by step solution

01

Understand the Function

The given function is \( y = (x^2 - x) e^{-x} \). We need to evaluate this function at \( x = 0 \).
02

Substitute \( x = 0 \) into the Function

Substitute \( x = 0 \) into the function: \( y = (0^2 - 0) e^{-0} \).
03

Simplify the Expression

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

\(4 \ln x=8\)

\(y=(x+\ln x)^{3}\) where \(x=1\)

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