The standard normal probability function is used to describe many different populations. Its graph is the well-known normal curve. This function is defined below. Give the intervals where the functions is increasing and decreasing. $$ f(x)=\frac{1}{\sqrt{5 \pi}} e^{-x^{2} / 4} $$

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We start with the function f(x) = (1/√(5π)) · e^(–x²/4). Since the constant (1/√(5π)) is positive, the behavior of f(x) is determined by the exponential part, e^(–x²/4). Step 1. Differentiate f(x): Let A = 1/√(5π), so f(x) = A · e^(–x²/4). Using the chain rule, compute the derivative of the exponential function: d/dx [e^(–x²/4)] = e^(–x²/4) · (d/dx (–x²/4)). Differentiate (–x²/4): d/dx (–x²/4) = –(1/2)x. Thus, the derivative becomes: f'(x) = A · e^(–x²/4) · (–(1/2)x) = –(A x)/(2) · e^(–x²/4). Step 2. Analyzing the sign of f'(x): Recall that: • A = 1/√(5π) > 0, • e^(–x²/4) > 0 for all x. Therefore, the sign of f'(x) depends only on –(x/2). • If x < 0, then –(x/2) is positive; hence f'(x) > 0. This implies f is increasing on the interval (–∞, 0). • If x > 0, then –(x/2) is negative; hence f'(x) < 0, which means f is decreasing on the interval (0, ∞). Step 3. Conclusion: The function f(x) is increasing for x in (–∞, 0) and decreasing for x in (0, ∞). The point x = 0 is a local maximum. Thus, the function increases on (–∞, 0) and decreases on (0, ∞). The function $$ f(x) = \frac{1}{\sqrt{5\pi}} e^{-x^{2}/4} $$ is increasing on the interval $$ (-\infty, 0) $$ and decreasing on the interval $$ (0, \infty) $$.

The standard normal probability function is used to describe many different populations. Its graph is the well-known normal curve. This function is defined below. Give the intervals where the functions is increasing and decreasing. \( f(x)=\frac{1}{\sqrt{5 \pi}} e^{-x^{2} / 4} \)

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