The value of `x, x in (2,oo)` where `f (x) = sqrt(x sqrt(8x-16))+ sqrt(x- sqrt(8x-16))` is not differentiable is:

Answer

To determine the value of \( x \) in the interval \( (2, \infty) \) where the function

\[ f(x) = \sqrt{x \sqrt{8x - 16}} + \sqrt{x - \sqrt{8x - 16}} \]

is not differentiable, we will follow these steps:

### Step 1: Identify the points where the function is not differentiable
The function \( f(x) \) involves square roots, which can become non-differentiable if the expression inside the square root becomes negative or zero. Therefore, we need to analyze the expressions inside the square roots.

### Step 2: Analyze the first square root
The first term is \( \sqrt{x \sqrt{8x - 16}} \). For this term to be defined, we need:

1. \( 8x - 16 \geq 0 \)

Solving this inequality:
\[ 8x \geq 16 \implies x \geq 2 \]

2. \( x \geq 0 \) (which is always true in the interval \( (2, \infty) \))

### Step 3: Analyze the second square root
The second term is \( \sqrt{x - \sqrt{8x - 16}} \). For this term to be defined, we need:

1. \( x - \sqrt{8x - 16} \geq 0 \)

Rearranging gives:
\[ x \geq \sqrt{8x - 16} \]

Squaring both sides (valid since both sides are non-negative in the interval \( (2, \infty) \)):
\[ x^2 \geq 8x - 16 \]
\[ x^2 - 8x + 16 \geq 0 \]
\[ (x - 4)^2 \geq 0 \]

This inequality holds for all \( x \), but it is equal to zero when \( x = 4 \).

### Step 4: Determine non-differentiable points
From the analysis, we see that \( f(x) \) is not differentiable at \( x = 4 \) because at this point, the expression \( x - \sqrt{8x - 16} \) becomes zero, leading to a potential cusp in the graph of the function.

### Step 5: Conclusion
Thus, the value of \( x \) in the interval \( (2, \infty) \) where \( f(x) \) is not differentiable is:

\[ \boxed{4} \]

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