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Magazine Chapter 4: Problem 22
Find the best possible bounds for the function. $$x^{3} e^{-x}, \text { for } x \geq 0$$
Short Answer
Expert verified
The bounds are from 0 to \( \frac{27}{e^3} \).
Step by step solution
01
Understand the Function
We are dealing with the function \( f(x) = x^3 e^{-x} \) where \( x \geq 0 \). Our task is to find its bounds, which means determining the maximum and minimum values of the function over its domain.
02
Find the Critical Points
To find the critical points, we first calculate the derivative of \( f(x) \). Using the product rule, the derivative is \( f'(x) = 3x^2 e^{-x} - x^3 e^{-x} = x^2 e^{-x} (3 - x) \). Set \( f'(x) = 0 \) to find critical points: \( x^2 e^{-x} (3 - x) = 0 \). This gives us \( x = 0 \) and \( x = 3 \) as critical points (since \( e^{-x} \) is never zero for \( x\geq 0 \)).
03
Evaluate at Critical Points and Endpoint
Evaluate \( f(x) \) at \( x = 0 \), which gives \( f(0) = 0^3 e^0 = 0 \). Next, evaluate \( f(x) \) at \( x = 3 \), which gives \( f(3) = 3^3 e^{-3} = \frac{27}{e^3} \). We also consider \( \lim_{x \to \infty} f(x) = \lim_{x \to \infty} x^3 e^{-x} = 0 \) since the exponential decay dominates.
04
Determine the Maximum Value
By evaluating the function at the critical points and considering the limit at infinity, we find that at \( x=3 \), the function \( f(x) = \frac{27}{e^3} \) achieves its maximum value over the interval \( [0, \infty) \). Therefore, the maximum bound of the function is \( \frac{27}{e^3} \).
05
Conclude the Bounds
The function is non-negative for \( x \geq 0 \) and tends to zero as \( x \to \infty \). The minimum value is found at \( x=0 \), which gives \( 0 \). Thus, the bounds are from \( 0 \) to \( \frac{27}{e^3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
Critical points play a crucial role in understanding the behavior of a function. They are points on the graph of a function where its derivative is zero or undefined.
For a smooth curve, these points often correspond to the peaks or troughs of the graph. In mathematical terms, critical points are locations where the slope of the tangent to the curve is zero, indicating a potential maximum, minimum, or saddle point.
For a smooth curve, these points often correspond to the peaks or troughs of the graph. In mathematical terms, critical points are locations where the slope of the tangent to the curve is zero, indicating a potential maximum, minimum, or saddle point.
- In this example, the function is given as \( f(x) = x^3 e^{-x} \).
- To find the critical points, compute the derivative and set it to zero.
- The derivative here is \( f'(x) = x^2 e^{-x} (3 - x) \), which gives critical points at \( x = 0 \) and \( x = 3 \).
Derivative
The derivative of a function provides information about the rate at which the function's value changes. It's a fundamental tool in calculus used to analyze and interpret function dynamics.
When calculating derivatives, the product rule is particularly useful for functions expressed as a product of two terms, such as \( f(x) = u(x)v(x) \). The product rule states that for such functions, the derivative is given by \( f'(x) = u'(x)v(x) + u(x)v'(x) \).
When calculating derivatives, the product rule is particularly useful for functions expressed as a product of two terms, such as \( f(x) = u(x)v(x) \). The product rule states that for such functions, the derivative is given by \( f'(x) = u'(x)v(x) + u(x)v'(x) \).
- For our function, apply the product rule to \( f(x) = x^3 e^{-x} \).
- Computing yields \( f'(x) = 3x^2 e^{-x} - x^3 e^{-x} = x^2 e^{-x} (3 - x) \).
Function Bounds
Function bounds describe the range of possible values a function can take on. These bounds are significant in determining how a function behaves within a specific interval.
In our example, the task is to find the highest (maximum) and lowest (minimum) values of the function \( f(x) = x^3 e^{-x} \) over the domain \( x \geq 0 \).
In our example, the task is to find the highest (maximum) and lowest (minimum) values of the function \( f(x) = x^3 e^{-x} \) over the domain \( x \geq 0 \).
- Evaluate the function at critical points found earlier, such as \( x = 0 \) and \( x = 3 \).
- Consider the behavior as \( x \) approaches infinity, where the exponential term suppresses the cubic growth, pushing the function towards zero.
- At \( x = 3 \), the function achieves its maximum value of \( \frac{27}{e^3} \).
- The minimum is at \( x = 0 \), giving \( 0 \).
Exponential Decay
Exponential decay is a process where a quantity decreases rapidly at a rate proportional to its current value. In our problem, the exponential term \( e^{-x} \) is the key player in determining the long-term behavior of the function \( f(x) = x^3 e^{-x} \).
Over time, as \( x \) increases, \( e^{-x} \) shrinks towards zero faster than the polynomial term \( x^3 \) grows, dominating the function's behavior.
Over time, as \( x \) increases, \( e^{-x} \) shrinks towards zero faster than the polynomial term \( x^3 \) grows, dominating the function's behavior.
- This is why the limit \( \lim_{x \to \infty} f(x) = 0 \); the exponential decay suppresses potential increases from \( x^3 \).
- The concept helps us understand how functions can approximate zero at large values, indicating a natural ceiling on the function's growth.
