6. Graphical Applications of Derivatives

The First Derivative Test

6. Graphical Applications of Derivatives

The First Derivative Test

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Multiple Choice

Identify the open intervals on which the function is increasing or decreasing.
$f(x)=xe^{-2x}$

Increasing on $\left(-\infty,\frac12\right)$ , Decreasing on $\left(\frac12,\infty\right)$

Increasing on $\left(\frac12,\infty\right)$ $\left(-\infty,\frac12\right)$ , Decreasing on $\left(-\infty,\frac12\right)$

Increasing on $\left(-\infty,\infty\right)$

Decreasing on $\left(-\infty,\infty\right)$

Verified step by step guidance

Step 1: To determine where the function f(x) = x * e^(-2x) is increasing or decreasing, start by finding its derivative f'(x). Use the product rule for differentiation, which states that if f(x) = u(x) * v(x), then f'(x) = u'(x) * v(x) + u(x) * v'(x). Here, u(x) = x and v(x) = e^(-2x).

Step 2: Compute the derivative of u(x) = x, which is u'(x) = 1. Then compute the derivative of v(x) = e^(-2x), which is v'(x) = -2 * e^(-2x) using the chain rule.

Step 3: Substitute the derivatives into the product rule formula: f'(x) = u'(x) * v(x) + u(x) * v'(x). This gives f'(x) = 1 * e^(-2x) + x * (-2 * e^(-2x)). Simplify the expression to f'(x) = e^(-2x) - 2x * e^(-2x).

Step 4: Factor out e^(-2x) from the derivative: f'(x) = e^(-2x) * (1 - 2x). Since e^(-2x) is always positive, the sign of f'(x) depends on the term (1 - 2x). Set 1 - 2x = 0 to find the critical point, which gives x = 1/2.

Step 5: Analyze the sign of f'(x) on the intervals determined by the critical point x = 1/2. For x < 1/2, (1 - 2x) > 0, so f'(x) > 0 and the function is increasing. For x > 1/2, (1 - 2x) < 0, so f'(x) < 0 and the function is decreasing. Thus, the function is increasing on (-∞, 1/2) and decreasing on (1/2, ∞).