Problem 63 - Algebra and Trigonometry

Question

62–63 ? Find, correct to two decimal places, (a) the intervals on which the function is increasing or decreasing, and (b) the range of the function. $$ y=x e^{-x} $$

Step-by-Step Solution

Answer

Increasing: \((-\infty, 1)\), Decreasing: \((1, \infty)\). Range: \((0, \frac{1}{e}]\).

Step 1: Find the derivative

To determine where the function is increasing or decreasing, start by finding the derivative of the function. The function given is \[ y = x e^{-x} \]We use the product rule, where the derivative of \( u imes v \) is \( u'v + uv' \). Here, let \( u = x \) and \( v = e^{-x} \).- Derivative of \( u = x \) is \( u' = 1 \).- Derivative of \( v = e^{-x} \) is \( v' = -e^{-x} \), using the chain rule.Applying the product rule, we get:\[ y' = 1 imes e^{-x} + x imes (-e^{-x}) = e^{-x} - x e^{-x} = (1-x)e^{-x} \]

Step 2: Set the derivative to zero

The critical points occur when the derivative is zero. Thus, set the derivative \( (1-x)e^{-x} \) equal to zero:\[ (1-x)e^{-x} = 0 \]Since \( e^{-x} \) is never zero, focus on \( 1-x = 0 \). Solving this gives \( x = 1 \). This is the critical point to test for increasing or decreasing intervals.

Step 3: Test intervals around the critical point

Test intervals around \( x = 1 \) to determine where the function is increasing or decreasing.- For \( x < 1 \), pick \( x = 0 \), then \( y'(0) = (1-0)e^{0} = 1 \), which is positive, indicating the function is increasing.- For \( x > 1 \), pick \( x = 2 \), then \( y'(2) = (1-2)e^{-2} = -\frac{1}{e^{2}} \), which is negative, indicating the function is decreasing.Thus, the function is increasing on \( (-\infty, 1) \) and decreasing on \( (1, \infty) \).

Step 4: Determine the range of the function

To find the range, evaluate the behavior of the function as \( x \rightarrow -\infty \) and \( x \rightarrow \infty \), along with its value at the critical point.- As \( x \rightarrow -\infty \), \( y = x e^{-x} \rightarrow 0 \) since \( e^{-x} \rightarrow \infty \) faster than \( x \rightarrow -\infty \).- As \( x \rightarrow \infty \), \( y = x e^{-x} \rightarrow 0 \) since \( e^{-x} \rightarrow 0 \) faster than \( x \rightarrow \infty \).- At \( x = 1 \), the function value is \[ y(1) = 1 e^{-1} = \frac{1}{e} \approx 0.3679 \].Thus, the range of the function is \( (0, \frac{1}{e}] \).

Step 5: Conclusion

The function \( y = x e^{-x} \) is increasing on \( (-\infty, 1) \) and decreasing on \( (1, \infty) \). The range of the function is \( (0, \frac{1}{e}] \).

Key Concepts

Derivative

In calculus, the derivative of a function measures how a function's output value changes as its input changes. Essentially, it provides the rate at which the function is increasing or decreasing at any given point. When we take the derivative of the function \( y = x e^{-x} \), we employ the product rule. This rule is essential because the function is the product of two functions: \( x \) and \( e^{-x} \).

According to the product rule, if you have two functions \( u \) and \( v \), the derivative of their product is \( u'v + uv' \).

• Here, if \( u = x \), then \( u' = 1 \).
• If \( v = e^{-x} \), the derivative \( v' \) requires us to use the chain rule, resulting in \( v' = -e^{-x} \).

Substituting back into the product rule gives us:\[ y' = 1 \times e^{-x} + x \times (-e^{-x}) = e^{-x} - x e^{-x} = (1-x)e^{-x} \]This derivative expression helps determine critical points and assess where the original function increases or decreases.

Increasing and Decreasing Functions

Once we have the derivative of a function, we can find where the function is increasing or decreasing. This involves identifying critical points where the derivative equals zero or changes sign. For the function \( y = x e^{-x} \), we set its derivative \((1-x)e^{-x}\) to zero:

• Since \( e^{-x} \) never equals zero, the factor \( (1-x) \) must be zero, giving us a critical point at \( x = 1 \).

Next, we test the intervals around this critical point:

• For \( x < 1 \), substituting values like \( x = 0 \), \( y'(0) = 1 \), shows the function is increasing.
• For \( x > 1 \), substituting values like \( x = 2 \), \( y'(2) = -\frac{1}{e^{2}} \), reveals the function is decreasing.

Thus, the function \( y = x e^{-x} \) is increasing on \((-\infty, 1)\) and decreasing on \((1, \infty)\). These regions reflect where the function slopes upward or downward.

Range of a Function

The range of a function specifies all possible output values a function can take as its input varies. For the function \( y = x e^{-x} \), finding the range involves looking at the behavior as \( x \) approaches extremes like \(-\infty\) and \(\infty\), as well as evaluating the function at the critical point.

• As \( x \rightarrow -\infty \), \( ye^{-x} \rightarrow 0 \); the exponential component dominates and grows faster than the negative \( x \).
• As \( x \rightarrow \infty \), \( y = x e^{-x} \rightarrow 0 \), since the exponential decays faster than linear growth.
• At the critical point \( x = 1 \), \( y = \frac{1}{e} \), which approximates to \( 0.3679 \).

Thus, the function's range is \( (0, \frac{1}{e}] \). This means \( y \) never hits zero but gets very close, spanning all values up to approximately 0.3679. Understanding the range helps you know what outputs you can expect to see given any valid input within the domain.