Questions: Consider the following. (If an answer does not exist, enter DNE.) F(x)=x sqrt(6-x) (a) Find the interval(s) of increase. (Enter your answer using interval Find the interval(s) of decrease. (Enter your answer using interval (b) Find the local minimum value(s). (Enter your answers as a comma- Find the local maximum value(s). (Enter your answers as a comma (c) Find the inflection point. (x, y)=( ) Find the interval(s) where the function is concave up. (Enter your answer using interval

Solution Steps

To solve the given problem, we need to analyze the function \( F(x) = x \sqrt{6-x} \) for its critical points, intervals of increase and decrease, local extrema, and concavity. Here's a high-level approach:

1. Find the derivative \( F'(x) \): This will help us determine the critical points and intervals of increase and decrease.
2. Critical Points: Set \( F'(x) = 0 \) to find critical points. These points will help identify potential local minima and maxima.
3. Intervals of Increase/Decrease: Use the sign of \( F'(x) \) to determine where the function is increasing or decreasing.
4. Second Derivative \( F''(x) \): This will help us find inflection points and determine concavity.
5. Inflection Points and Concavity: Set \( F''(x) = 0 \) to find inflection points and use the sign of \( F''(x) \) to determine intervals of concavity.

We start with the function \( F(x) = x \sqrt{6 - x} \). The first derivative is calculated as follows: \[ F'(x) = -\frac{x}{2\sqrt{6 - x}} + \sqrt{6 - x} \]

Setting the first derivative equal to zero to find critical points: \[ F'(x) = 0 \implies -\frac{x}{2\sqrt{6 - x}} + \sqrt{6 - x} = 0 \] This gives us the critical point: \[ x = 4 \]

To find the intervals of increase and decrease, we analyze the sign of \( F'(x) \):

• For \( x < 4 \), \( F'(x) > 0 \) (function is increasing).
• For \( x > 4 \), \( F'(x) < 0 \) (function is decreasing).

Thus, we conclude:

• Intervals of Increase: \( (-\infty, 4) \)
• Intervals of Decrease: \( (4, \infty) \)

The second derivative is given by: \[ F''(x) = -\frac{x}{4(6 - x)^{3/2}} - \frac{1}{\sqrt{6 - x}} \]

Setting the second derivative equal to zero to find inflection points: \[ F''(x) = 0 \implies -\frac{x}{4(6 - x)^{3/2}} - \frac{1}{\sqrt{6 - x}} = 0 \] This gives us the inflection point: \[ x = 8 \]

To find the intervals of concavity, we analyze the sign of \( F''(x) \):

• For \( x < 8 \), \( F''(x) < 0 \) (function is concave down).
• For \( x > 8 \), \( F''(x) > 0 \) (function is concave up).

Thus, we conclude:

• Intervals Concave Up: \( (8, \infty) \)
• Intervals Concave Down: \( (-\infty, 8) \)

Final Answer