Find the points of local maxima and local minima, if any, of the following functions. Find also the local maximum and local minimum values :
`f(x)=xsqrt(1-x), x gt0`.

Answer

To find the points of local maxima and minima for the function \( f(x) = x \sqrt{1 - x} \) where \( x > 0 \), we will follow these steps:

### Step 1: Find the derivative of the function
To find the critical points, we first need to compute the derivative \( f'(x) \).

Using the product rule:
\[ f(x) = x \cdot (1 - x)^{1/2} \]
Let \( u = x \) and \( v = (1 - x)^{1/2} \). Then,
\[ f'(x) = u'v + uv' \]
where \( u' = 1 \) and \( v' = \frac{1}{2}(1 - x)^{-1/2}(-1) = -\frac{1}{2\sqrt{1 - x}} \).

Now substituting these into the product rule:
\[ f'(x) = 1 \cdot (1 - x)^{1/2} + x \cdot \left(-\frac{1}{2\sqrt{1 - x}}\right) \]
\[ = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set \( f'(x) = 0 \):
\[ \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} = 0 \]
Multiplying through by \( 2\sqrt{1 - x} \) to eliminate the fraction:
\[ 2(1 - x) - x = 0 \]
\[ 2 - 2x - x = 0 \implies 2 - 3x = 0 \implies 3x = 2 \implies x = \frac{2}{3} \]

### Step 3: Determine if it is a maximum or minimum
Next, we need to find the second derivative \( f''(x) \) to determine whether this critical point is a maximum or minimum.

Using the quotient rule on \( f'(x) \):
\[ f'(x) = \frac{2(1 - x) - x}{2\sqrt{1 - x}} = \frac{2 - 3x}{2\sqrt{1 - x}} \]
Now, we differentiate this again:
Let \( u = 2 - 3x \) and \( v = 2\sqrt{1 - x} \):
\[ f''(x) = \frac{u'v - uv'}{v^2} \]
Calculating \( u' = -3 \) and \( v' = \frac{1}{2}(1 - x)^{-1/2}(-1) = -\frac{1}{2\sqrt{1 - x}} \):
\[ f''(x) = \frac{(-3)(2\sqrt{1 - x}) - (2 - 3x)\left(-\frac{1}{2\sqrt{1 - x}}\right)}{(2\sqrt{1 - x})^2} \]
Substituting \( x = \frac{2}{3} \):
\[ f''\left(\frac{2}{3}\right) = \frac{-6\sqrt{1 - \frac{2}{3}} + (2 - 3 \cdot \frac{2}{3})\left(-\frac{1}{2\sqrt{1 - \frac{2}{3}}}\right)}{(2\sqrt{1 - \frac{2}{3}})^2} \]
Calculating \( 1 - \frac{2}{3} = \frac{1}{3} \):
\[ = \frac{-6\sqrt{\frac{1}{3}} + 0}{(2\sqrt{\frac{1}{3}})^2} \]
Since \( f''\left(\frac{2}{3}\right) < 0 \), it indicates a local maximum.

### Step 4: Find the maximum value
Now we calculate the maximum value of \( f \) at \( x = \frac{2}{3} \):
\[ f\left(\frac{2}{3}\right) = \frac{2}{3}\sqrt{1 - \frac{2}{3}} = \frac{2}{3}\sqrt{\frac{1}{3}} = \frac{2}{3} \cdot \frac{1}{\sqrt{3}} = \frac{2}{3\sqrt{3}} = \frac{2\sqrt{3}}{9} \]

### Final Result
- Local maximum point: \( x = \frac{2}{3} \)
- Local maximum value: \( f\left(\frac{2}{3}\right) = \frac{2\sqrt{3}}{9} \)
- There are no local minima in the given domain \( x > 0 \).