Find the Critical Points y=x^2e^(-x)
The problem is to determine the critical points of the function y = x^2e^(-x). A critical point of a function is a point at which its derivative is either zero or undefined. To find the critical points of a given function, you first need to take the derivative of the function with respect to x, and then solve for the values of x that make this derivative equal to zero or lead to an undefined derivative. The found values for x correspond to the critical points of the function on its domain.
$y = x^{2} e^{- x}$
Answer
Solution:
Step:1
Find the first derivative.
Step:1.1
Calculate the derivative of the function.
Step:1.1.1
Apply the Product Rule: $\frac{d}{dx}(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$, where $f(x) = x^2$ and $g(x) = e^{-x}$.
Step:1.1.2
Use the Chain Rule: $\frac{d}{dx}(f(g(x))) = f'(g(x))g'(x)$, where $f(u) = e^u$ and $g(x) = -x$.
Step:1.1.2.1
Let $u = -x$. Compute $\frac{d}{du}(e^u)\frac{d}{dx}(-x)$.
Step:1.1.2.2
Apply the Exponential Rule: $\frac{d}{du}(a^u) = a^u \ln(a)$, where $a = e$.
Step:1.1.2.3
Replace $u$ with $-x$ to get $x^2(e^{-x} \frac{d}{dx}(-x)) + e^{-x} \frac{d}{dx}(x^2)$.
Step:1.1.3
Differentiate the terms.
Step:1.1.3.1
The derivative of $-x$ is $-1$ since it's a linear term.
Step:1.1.3.2
Apply the Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$, where $n = 1$.
Step:1.1.3.3
Simplify the derivative expression.
Step:1.1.3.3.1
Multiply $-1$ by $1$.
Step:1.1.3.3.2
Rearrange the terms to get $-e^{-x}x^2 + e^{-x}2x$.
Step:1.1.3.3.3
Rewrite $-1e^{-x}$ as $-e^{-x}$.
Step:1.1.3.4
Apply the Power Rule again for $n = 2$.
Step:1.1.4
Combine terms to get $f'(x) = -x^2e^{-x} + 2xe^{-x}$.
Step:1.2
The derivative $f'(x)$ is $-x^2e^{-x} + 2xe^{-x}$.
Step:2
Solve the equation $f'(x) = 0$.
Step:2.1
Set the derivative equal to zero.
Step:2.2
Factor out the common term $xe^{-x}$.
Step:2.2.1
Factor $xe^{-x}$ from $-x^2e^{-x}$.
Step:2.2.2
Factor $xe^{-x}$ from $2xe^{-x}$.
Step:2.2.3
Factor out $xe^{-x}$ from the entire expression.
Step:2.3
Determine the values that make each factor zero.
Step:2.4
Solve for $x$ in each factor.
Step:2.5
There is no solution for $e^{-x} = 0$ as the exponential function never equals zero.
Step:2.6
Solve $-x + 2 = 0$ for $x$.
Step:2.7
The critical points are $x = 0$ and $x = 2$.
Step:3
There are no points where the derivative is undefined.
Step:4
Evaluate the original function at the critical points.
Step:4.1
Calculate the function value at $x = 0$.
Step:4.2
Calculate the function value at $x = 2$.
Step:4.3
List all critical points: $(0, 0)$ and $(2, \frac{4}{e^2})$.
Knowledge Notes:
Product Rule: When differentiating the product of two functions $f(x)$ and $g(x)$, the derivative is $f'(x)g(x) + f(x)g'(x)$.
Chain Rule: Used to differentiate a composite function. If $y = f(g(x))$, then the derivative is $f'(g(x))g'(x)$.
Exponential Rule: The derivative of $a^u$, where $a$ is a constant and $u$ is a function of $x$, is $a^u \ln(a) \cdot u'$.
Power Rule: For any real number $n$, the derivative of $x^n$ with respect to $x$ is $nx^{n-1}$.
Critical Points: Points on the graph of a function where the derivative is zero or undefined. These points are potential locations for local maxima, minima, or points of inflection.
Exponential Functions: Functions of the form $a^x$ where $a$ is a positive constant. The exponential function $e^x$ is particularly important due to its unique properties in calculus.
Natural Logarithm: The function $\ln(x)$ is the inverse of the exponential function $e^x$. It is undefined for $x \leq 0$.
Factoring: The process of expressing an expression as a product of its factors. This is useful for solving equations set to zero, as it allows us to apply the Zero Product Property.
Zero Product Property: If a product of factors equals zero, at least one of the factors must be zero. This property is used to find the roots of polynomial equations.
