Problem 40 - Calculus A New Horizon

Question

(a) Find the limits of the function as \(x \rightarrow+\infty\) and \(x \rightarrow-\infty .\) (b) Give a complete graph of the function, and identify the location of all relative extrema and inflection points. Check your work with a graphing utility. $$x e^{-2 x}$$

Step-by-Step Solution

Answer

The limits are 0 as x → +∞ and -∞ as x → -∞. Relative maximum and inflection at x = 0.5.

Step 1: Analyzing the Function as x Approaches Infinity

To find the limit of the function \( f(x) = x e^{-2x} \) as \( x \rightarrow +\infty \), observe that the exponential term \( e^{-2x} \) approaches zero. So, \( x \cdot e^{-2x} \) becomes \( x \cdot 0 \), which approaches zero. Therefore, the limit is \( \lim_{{x \to +\infty}} x e^{-2x} = 0 \).

Step 2: Analyzing the Function as x Approaches Negative Infinity

As \( x \rightarrow -\infty \), \( e^{-2x} \rightarrow +\infty \) because \( e^{-2x} \) behaves like \( e^{2|x|} \). Therefore, the product \( x \cdot e^{-2x} \) behaves like \( -\infty \times +\infty \), leading to negative infinity. Hence, the limit is \( \lim_{{x \to -\infty}} x e^{-2x} = -\infty \).

Step 3: Finding Critical Points for Relative Extrema

To find critical points, take the derivative \( f'(x) \) and set it to zero. First, find \( f'(x) \) using the product rule: \( f'(x) = e^{-2x} + x(-2e^{-2x}) = e^{-2x} - 2xe^{-2x} \). Set \( f'(x) = 0 \), leading to \( e^{-2x}(1 - 2x) = 0 \). The solution is \( x = 0.5 \).

Step 4: Determine Nature of Critical Points

Examine \( f'(x) \) around \( x = 0.5 \) to determine the type of extrema. Evaluate \( f'(x) \) slightly before and after \( x = 0.5 \). For \( x < 0.5 \), \( f'(x) \) is positive, and for \( x > 0.5 \), \( f'(x) \) is negative. Hence, \( x = 0.5 \) is a local maximum.

Step 5: Identifying Inflection Points

To find inflection points, examine the second derivative \( f''(x) \). Compute \( f''(x) = \frac{d}{dx}(e^{-2x} - 2xe^{-2x}) = -2e^{-2x} - 2(-2e^{-2x} - 2xe^{-2x}) = -2e^{-2x} + 4e^{-2x} - 4xe^{-2x} = 2e^{-2x} - 4xe^{-2x} \). Set \( f''(x) = 0 \): \( 2e^{-2x}(1 - 2x) = 0 \), giving \( x = 0.5 \). \( x = 0.5 \) is also an inflection point where concavity changes.

Step 6: Sketch the Graph

Consider the key points: limits at \( \pm\infty \), the maximum at \( x = 0.5 \), and the inflection point at \( x = 0.5 \). The graph should start from bottom-left, rise to the maximum at \( x = 0.5 \), and then decrease, approaching zero along the positive x-axis.

Key Concepts

Limits

When studying functions, limits help us understand the behavior of a function as the input (in this case, x) moves towards infinity or negative infinity. Limits tell us what value a function approaches as the input grows very large or very small. For the function given, \( f(x) = x e^{-2x} \), as \( x \rightarrow +finity \), the exponential term \( e^{-2x} \) approaches zero. Thus, the limit is zero because multiplying by x becomes insignificant. As \( x \rightarrow -finity \), \( e^{-2x} \) becomes very large, and since x is negative here, the product \( x e^{-2x} \) trends towards negative infinity.

Critical Points

Critical points are where the derivative of a function equals zero or is undefined. These points help identify locations of possible maxima, minima, or saddle points of the function. Calculating the derivative, \( f'(x) = e^{-2x} - 2x e^{-2x} \), and setting it to zero, we solve: \( e^{-2x}(1 - 2x) = 0 \). This leads us to the critical point at \( x = 0.5 \). To determine the nature of this point (whether it's a maxima or minima), we evaluate the sign of the derivative around this point. When \( x < 0.5 \), \( f'(x) > 0 \) indicating increasing function, and when \( x > 0.5 \), \( f'(x) < 0 \) indicating decreasing function. This confirms that \( x = 0.5 \) is a local maximum.

Inflection Points

Inflection points are where a function's concavity changes. To find these, the second derivative of the function is used. For the given function, the second derivative is \( f''(x) = 2e^{-2x} - 4xe^{-2x} \). Setting the second derivative equal to zero, \( 2e^{-2x}(1 - 2x) = 0 \), we find an inflection point at \( x = 0.5 \). This indicates a switch in concavity at this point. Around \( x = 0.5 \), checking the second derivative sign confirms a transition, confirming \( x = 0.5 \) as an inflection point.

Derivative

Derivatives are the rate of change of a function with respect to a variable, often providing the slope of the function at any given point. In practical terms, the derivative helps to understand how a function behaves as its inputs change. For our function \( f(x) = x e^{-2x} \), the derivative, \( f'(x) = e^{-2x} - 2x e^{-2x} \) using the product rule, helps find critical points by identifying where the slope is zero. The second derivative, \( f''(x) = 2e^{-2x} - 4xe^{-2x} \), provides insight into changes in concavity, assisting in locating inflection points. Understanding derivatives, therefore, allows a deeper analysis of a function's shape and behavior.