Find the range of the following functions:
`y=sqrt(6-x)+2sqrt(x-4)`

Answer

To find the range of the function \( y = \sqrt{6 - x} + 2\sqrt{x - 4} \), we will follow these steps:

### Step 1: Determine the domain of the function
The function involves square roots, which require the expressions inside them to be non-negative.

1. For \( \sqrt{6 - x} \):
\[ 6 - x \geq 0 \implies x \leq 6 \]

2. For \( \sqrt{x - 4} \):
\[ x - 4 \geq 0 \implies x \geq 4 \]

Combining these inequalities, we find that the domain of \( y \) is:
\[ 4 \leq x \leq 6 \]

### Step 2: Analyze the function within the domain
Next, we need to evaluate the function at the endpoints of the domain and find any critical points within the interval.

### Step 3: Calculate the function values at the endpoints
1. At \( x = 4 \):
\[ y(4) = \sqrt{6 - 4} + 2\sqrt{4 - 4} = \sqrt{2} + 2 \cdot 0 = \sqrt{2} \]

2. At \( x = 6 \):
\[ y(6) = \sqrt{6 - 6} + 2\sqrt{6 - 4} = 0 + 2\sqrt{2} = 2\sqrt{2} \]

### Step 4: Find critical points by differentiating
To find any local maxima or minima, we differentiate \( y \) with respect to \( x \):
\[ y = \sqrt{6 - x} + 2\sqrt{x - 4} \]
Differentiating:
\[ \frac{dy}{dx} = \frac{-1}{2\sqrt{6 - x}} + \frac{2}{2\sqrt{x - 4}} = -\frac{1}{2\sqrt{6 - x}} + \frac{1}{\sqrt{x - 4}} \]
Setting the derivative to zero to find critical points:
\[ -\frac{1}{2\sqrt{6 - x}} + \frac{1}{\sqrt{x - 4}} = 0 \]
This implies:
\[ \frac{1}{\sqrt{x - 4}} = \frac{1}{2\sqrt{6 - x}} \implies 2\sqrt{6 - x} = \sqrt{x - 4} \]
Squaring both sides:
\[ 4(6 - x) = x - 4 \implies 24 - 4x = x - 4 \implies 5x = 28 \implies x = \frac{28}{5} = 5.6 \]

### Step 5: Calculate the function value at the critical point
Now, we evaluate \( y \) at \( x = 5.6 \):
\[ y(5.6) = \sqrt{6 - 5.6} + 2\sqrt{5.6 - 4} = \sqrt{0.4} + 2\sqrt{1.6} \]
Calculating these values:
\[ \sqrt{0.4} = \frac{2}{\sqrt{10}} \quad \text{and} \quad 2\sqrt{1.6} = 2 \cdot \frac{4}{\sqrt{10}} = \frac{8}{\sqrt{10}} \]
Thus,
\[ y(5.6) = \frac{2 + 8}{\sqrt{10}} = \frac{10}{\sqrt{10}} = \sqrt{10} \]

### Step 6: Determine the range
Now we have the values:
- At \( x = 4 \), \( y(4) = \sqrt{2} \)
- At \( x = 5.6 \), \( y(5.6) = \sqrt{10} \)
- At \( x = 6 \), \( y(6) = 2\sqrt{2} \)

The minimum value is \( \sqrt{2} \) and the maximum value is \( 2\sqrt{2} \).

### Final Result
Thus, the range of the function \( y = \sqrt{6 - x} + 2\sqrt{x - 4} \) is:
\[ \text{Range} = [\sqrt{2}, 2\sqrt{2}] \]

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