Problem 80 - Calculus

Question

Are the statements true or false? Give an explanation for your answer. The function \(f(x)=e^{-x^{2}}\) is decreasing for all \(x\).

Step-by-Step Solution

Answer

The statement is false; the function increases for \( x < 0 \) and decreases for \( x > 0 \).

Step 1: Understand the Function

The function given is \( f(x) = e^{-x^2} \), which is an exponential function where the exponent is a quadratic function \(-x^2\).

Step 2: Find the Derivative

To determine where the function is increasing or decreasing, we need to find the first derivative of \( f(x) \). For \( f(x) = e^{-x^2} \), use the chain rule: \( f'(x) = \frac{d}{dx} (e^{-x^2}) = e^{-x^2} \cdot (-2x) = -2xe^{-x^2} \).

Step 3: Analyze the Sign of the Derivative

The sign of \( f'(x) = -2xe^{-x^2} \) determines where \( f(x) \) is increasing or decreasing. Since \( e^{-x^2} \) is always positive for all \( x \) because it's an exponential function, the sign is determined by \(-2x\).

Step 4: Determine Increasing/Decreasing Intervals

The term \(-2x\) changes sign at \( x = 0 \). For \( x > 0 \), \(-2x < 0\), so \( f'(x) < 0 \) and the function is decreasing. For \( x < 0 \), \(-2x > 0\), so \( f'(x) > 0 \) and the function is increasing.

Step 5: Conclude from Analysis

Since the derivative \( f'(x) \) is positive for \( x < 0 \) and negative for \( x > 0 \), \( f(x) = e^{-x^2} \) is not decreasing for all \( x \). It is increasing for \( x < 0 \) and decreasing for \( x > 0 \).

Key Concepts

Exponential Function

The function given in the exercise, \( f(x) = e^{-x^2} \), is a classic example of an exponential function where the exponent is itself a function. In simpler terms, an exponential function generally takes the form \(f(x) = a^{g(x)}\) where \(a\) is a constant. Here, \(e\) is the base of the natural logarithm, approximately 2.718, and it's important because exponential functions with this base have properties useful for calculus. Unlike polynomials, exponential functions show rapid growth or decay depending on the sign of the exponent. In this exercise, the exponent \(-x^2\) makes the function show a dampening effect, and the values decrease as \(x\) moves away from zero in either direction.

Chain Rule

To find the derivative of functions like \(f(x) = e^{-x^2}\), we employ the chain rule. This rule is particularly useful when dealing with composite functions, which are functions nested inside one another, as is the case here: an exponential function with a quadratic function in its exponent. The chain rule is written as:

• If \(h(x) = f(g(x))\), then \(h'(x) = f'(g(x)) \cdot g'(x)\).

In our function, let \(g(x) = -x^2\), then the derivative \(g'(x) = -2x\). Therefore, applying the chain rule results in \(f'(x) = e^{-x^2} \cdot (-2x)\), simplifying to \(-2xe^{-x^2}\). This result shows us how the slopes and turning points of the function behave.

First Derivative

Calculating the first derivative of a function is crucial in understanding its behavior over its domain. The first derivative, denoted \(f'(x)\), shows the rate of change of the function with respect to changes in \(x\). Simply put, it lets us know if the function is going up or down at any point. For our function \(f(x) = e^{-x^2}\), we found that the first derivative is \(f'(x) = -2xe^{-x^2}\). Since the exponential term \(e^{-x^2}\) is always positive, the sign of \(f'(x)\) is determined by \(-2x\). Thus, when \(x > 0\), \(f'(x)\) is negative, indicating that the function is decreasing. Conversely, when \(x < 0\), \(f'(x)\) is positive, and the function is increasing.

Increasing and Decreasing Functions

Analyzing whether a function is increasing or decreasing based on its first derivative is a key part of calculus. By studying \(f'(x)\), we can describe the behavior of \(f(x)\) in different intervals. In our exercise, since \(f'(x) = -2xe^{-x^2}\), and \(e^{-x^2}\) never changes sign, the determining factor is \(-2x\). Based on this:

• For \(x > 0\), \(-2x\) is negative, so \(f(x)\) is decreasing.
• For \(x < 0\), \(-2x\) is positive, so \(f(x)\) is increasing.
• At \(x = 0\), \(f'(x) = 0\), indicating a critical point.

Thus, \(f(x) = e^{-x^2}\) increases on \((-\infty, 0)\) and decreases on \((0, \infty)\). This analysis helps us understand the graph's nature, showing a peak at zero.