Problem 35 - Single Variable Calculus

Question

(a) find the intervals on which \(f\) is increasing or decreasing, and (b) find the relative maxima and relative minima of \(\vec{f}\). $$ f(x)=x^{2} e^{-x} $$

Step-by-Step Solution

Answer

(a) Increasing on the interval \((0, 2)\), decreasing on the intervals \((-\infty, 0)\) and \((2, \infty)\). (b) Relative minimum at \((0,0)\) and relative maximum at \((2, 4e^{-2})\).

Step 1: Find the first derivative of the function

To find the first derivative of the function \(f(x)\), we will use the product rule, which states that if a function is given by the product of two functions following the form \(f(x) = g(x)h(x)\), then the derivative of \(f(x)\) can be found by differentiating one of the components and multiplying by the non-differentiated component, then switching the roles of the components, and finally adding the results together. In this case, \(g(x) = x^2\) and \(h(x) = e^{-x}\). Differentiate the function using the product rule: \(f'(x) = (2x)e^{-x} + x^2(-e^{-x})\) Simplify the first derivative: \(f'(x) = e^{-x}(2x - x^2)\)

Step 2: Find the critical points

A critical point occurs when the first derivative is equal to 0 or is undefined. In this case, the first derivative is always defined, so we will find the values of x where the first derivative is equal to 0. \(f'(x) = e^{-x}(2x - x^2) = 0\) e^{-x} is always positive, so we only need to find the roots of the polynomial part: \(2x - x^2 = 0\) Factor out a common term of x: \(x(2 - x) = 0\) Thus, we have two critical points: \(x=0\) and \(x=2\).

Step 3: Determine increasing and decreasing intervals

To determine the intervals on which the function is increasing or decreasing, analyze the sign of \(f'(x)\) for the intervals separated by the critical points. Test the intervals \(x<0\), \(02\). For \(x < 0\): \(f'(x) = e^{-x}(2x - x^2)\). Since \(e^{-x} > 0\) and \(2x - x^2 < 0\), \(f'(x) < 0\). Therefore, the function is decreasing on the interval \((-\infty, 0)\). For \(0 < x < 2\): \(f'(x) = e^{-x}(2x - x^2)\). Since \(e^{-x} > 0\) and \(2x - x^2 > 0\), \(f'(x) > 0\). Therefore, the function is increasing on the interval \((0, 2)\). For \(x > 2\): \(f'(x) = e^{-x}(2x - x^2)\). Since \(e^{-x} > 0\) and \(2x - x^2 < 0\), \(f'(x) < 0\). Therefore, the function is decreasing on the interval \((2, \infty)\).

Step 4: Find relative maxima and minima

We have two critical points, \(x=0\) and \(x=2\). To determine if they are relative maxima or minima, examine the change in the sign of the first derivative right before and after the critical points. Alternatively, you can use the second derivative test. For \(x=0\), we found that the sign of the first derivative changed from negative to positive, which means there is a relative minimum at this point. Therefore, we have a relative minimum at \((0,f(0)) = (0,0)\). For \(x=2\), we found that the sign of the first derivative changed from positive to negative, which means there is a relative maximum at this point. Therefore, we have a relative maximum at \((2,f(2)) = (2, 4e^{-2})\). Thus, the solution is: (a) The function is increasing on the interval \((0, 2)\) and decreasing on the intervals \((-\infty, 0)\) and \((2, \infty)\). (b) There is a relative minimum at \((0,0)\) and a relative maximum at \((2, 4e^{-2})\).

Key Concepts

Critical Points

Critical points are fundamental in calculus as they help identify where a function changes its increasing or decreasing behavior. Critical points occur where the first derivative of a function equals zero or is undefined. In our example function, \( f(x) = x^2 e^{-x} \), finding the critical points involves determining where the derivative, \( f'(x) = e^{-x}(2x - x^2) \), becomes zero. For this derivative, \( e^{-x} \) is always positive; hence we only need to focus on the polynomial part, \( 2x - x^2 = 0 \). Solving this gives us the critical points \( x = 0 \) and \( x = 2 \). These points are crucial for further analysis, like determining intervals of increase or decrease, and finding relative maxima or minima.

Increasing Decreasing Functions

To determine where a function is increasing or decreasing, we analyze the sign of its first derivative. The sign of \( f'(x) \) tells us about the behavior of the function \( f(x) \):

• If \( f'(x) > 0 \), the function is increasing on that interval.
• If \( f'(x) < 0 \), the function is decreasing.

We check the signs of \( f'(x) = e^{-x}(2x - x^2) \) across different intervals defined by the critical points \( x = 0 \) and \( x = 2 \).
For \( x < 0 \), \( f'(x) < 0 \), meaning the function decreases. Between \( 0 < x < 2 \), \( f'(x) > 0 \), so the function increases. Lastly, for \( x > 2 \), \( f'(x) < 0 \) again, indicating a decrease. This helps us understand where the function climbs or descends across its domain.

First Derivative Test

The first derivative test is a handy tool to determine the nature of relative extrema (peaks or valleys) at critical points. When you have a critical point at \( x = c \), check the sign of \( f'(x) \) before and after \( c \) to understand the function's behavior:

• If \( f'(x) \) changes from positive to negative at \( c \), there's a relative maximum at \( c \).
• If \( f'(x) \) changes from negative to positive at \( c \), there's a relative minimum.
• If \( f'(x) \) does not change sign, no extremum exists.

For our function \( f(x) = x^2 e^{-x} \), at the critical point \( x = 0 \), \( f'(x) \) changes from negative to positive, indicating a relative minimum at \( (0, 0) \). For \( x = 2 \), \( f'(x) \) changes from positive to negative, indicating a relative maximum at \( (2, 4e^{-2}) \). This method effectively uncovers the nature of each critical point.

Relative Maxima and Minima

Relative maxima and minima are points where a function reaches local peaks or troughs. These are determined using critical points and analyzing the function's behavior in their neighborhood. For \( f(x) = x^2 e^{-x} \), we found the critical points to be \( x = 0 \) and \( x = 2 \).
Examining these points using the first derivative test:

• At \( x = 0 \): The derivative \( f'(x) \) changes from negative to positive, indicating a relative minimum with value \( f(0) = 0 \).
• At \( x = 2 \): The derivative \( f'(x) \) changes from positive to negative, indicating a relative maximum with value \( f(2) = 4e^{-2} \).

These extrema highlight where the function achieves localized low and high values, crucial for understanding the graph's behavior.