The minimum value of the function
`f(x)=x^(3//2)+x^(-3//2)-4(x+1/x)` for all permissible real x is

Options

A. `-10`
B. `-6`
C. `-7`
D. `-8`

Explanation

To find the minimum value of the function \( f(x) = x^{3/2} + x^{-3/2} - 4\left(x + \frac{1}{x}\right) \), we can follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = x^{3/2} + x^{-3/2} - 4\left(x + \frac{1}{x}\right) \]

### Step 2: Substitute \( t = \sqrt{x} + \frac{1}{\sqrt{x}} \)
Notice that:
\[ \sqrt{x} + \frac{1}{\sqrt{x}} \geq 2 \]
This is due to the AM-GM inequality. Let \( t = \sqrt{x} + \frac{1}{\sqrt{x}} \). Then:
\[ x^{3/2} + x^{-3/2} = (\sqrt{x})^3 + \left(\frac{1}{\sqrt{x}}\right)^3 = t^3 - 3t \]
Thus, we can rewrite \( f(x) \) as:
\[ f(x) = (t^3 - 3t) - 4t = t^3 - 4t - 3t = t^3 - 7t \]

### Step 3: Find the derivative
Now we differentiate \( f(t) \):
\[ f'(t) = 3t^2 - 7 \]

### Step 4: Set the derivative to zero
To find the critical points, set the derivative equal to zero:
\[ 3t^2 - 7 = 0 \]
Solving for \( t \):
\[ 3t^2 = 7 \implies t^2 = \frac{7}{3} \implies t = \sqrt{\frac{7}{3}} \text{ (only positive root is valid since \( t \geq 2 \))} \]

### Step 5: Check if \( t \) is permissible
We need to check if \( t = \sqrt{\frac{7}{3}} \) is greater than or equal to 2:
\[ \sqrt{\frac{7}{3}} \approx 1.53 \text{ (which is less than 2)} \]
Thus, this critical point is not permissible.

### Step 6: Evaluate \( f(t) \) at the boundary
Since \( t \geq 2 \), we evaluate \( f(t) \) at \( t = 2 \):
\[ f(2) = 2^3 - 7 \cdot 2 = 8 - 14 = -6 \]

### Step 7: Check the behavior as \( t \to \infty \)
As \( t \) increases, \( f(t) \) behaves like \( t^3 \) which goes to \( +\infty \). Therefore, the minimum value occurs at \( t = 2 \).

### Conclusion
The minimum value of the function \( f(x) \) is:
\[ \boxed{-6} \]