Questions: Find the open intervals on which the function is increasing or decreasing. h(x)=9 sqrt(x) e^(-x) increasing □ decreasing □

Transcript text: Find the open intervals on which the function is increasing or decreasing. \[ h(x)=9 \sqrt{x} \mathrm{e}^{-x} \] increasing $\square$ decreasing $\square$

Solution

Solution Steps

To determine where the function is increasing or decreasing, we need to find the derivative of the function and analyze its sign. The function is increasing where the derivative is positive and decreasing where it is negative.

Step 1: Find the Derivative

To determine where the function $ h(x) = 9 \sqrt{x} e^{-x} $ is increasing or decreasing, we first find its derivative: \[ h'(x) = -9 \sqrt{x} e^{-x} + \frac{9 e^{-x}}{2 \sqrt{x}} \]

Step 2: Find Critical Points

Set the derivative equal to zero to find critical points: \[ -9 \sqrt{x} e^{-x} + \frac{9 e^{-x}}{2 \sqrt{x}} = 0 \] Solving this equation, we find the critical point: \[ x = \frac{1}{2} \]

Step 3: Determine Intervals of Increase and Decrease

Analyze the sign of $ h'(x) $ in the intervals defined by the critical point $ x = \frac{1}{2} $.

For $ x \in (0, \frac{1}{2}) $, $ h'(x) > 0 $, so the function is increasing.
For $ x \in (\frac{1}{2}, \infty) $, $ h'(x) < 0 $, so the function is decreasing.