Question:

Step 1: Determine the domain of the function. For the function $f(x) = x^{\frac{3}{2}} + x^{\frac{-3}{2}} - 4\left(x + \dfrac{1}{x}\right)$ to be defined, we need $x > 0$ since fractional powers with negative exponents and the term $\dfrac{1}{x}$ require $x$ to be positive. Step 2: Rewrite the function in a more convenient form. We can express the function as: $$f(x) = x^{\frac{3}{2}} + \dfrac{1}{x^{\frac{3}{2}}} - 4\left(x + \dfrac{1}{x}\right)$$ Step 3: Apply the AM-GM inequality to the first pair of terms. For positive numbers, the AM-GM inequality states that the arithmetic mean is greater than or equal to the geometric mean. Applying this to $x^{\frac{3}{2}}$ and $x^{\frac{-3}{2}}$: $$x^{\frac{3}{2}} + x^{\frac{-3}{2}} \geq 2\sqrt{x^{\frac{3}{2}} \cdot x^{\frac{-3}{2}}} = 2\sqrt{x^0} = 2$$ Equality holds when $x^{\frac{3}{2}} = x^{\frac{-3}{2}}$, which occurs when $x = 1$. Step 4: Apply the AM-GM inequality to the second pair of terms. Similarly, for $x$ and $\dfrac{1}{x}$: $$x + \dfrac{1}{x} \geq 2\sqrt{x \cdot \dfrac{1}{x}} = 2\sqrt{1} = 2$$ Equality holds when $x = \dfrac{1}{x}$, which also occurs when $x = 1$. Step 5: Combine the inequalities to find a lower bound for $f(x)$. From the AM-GM results: $$f(x) = x^{\frac{3}{2}} + x^{\frac{-3}{2}} - 4\left(x + \dfrac{1}{x}\right) \geq 2 - 4(2) = 2 - 8 = -6$$ Step 6: Verify that the minimum is achieved at $x = 1$. Since both AM-GM equalities are satisfied when $x = 1$, we can verify by direct substitution: $$f(1) = 1^{\frac{3}{2}} + 1^{\frac{-3}{2}} - 4\left(1 + \dfrac{1}{1}\right) = 1 + 1 - 4(2) = 2 - 8 = -6$$ Step 7: State the final answer. The minimum value of the function $f(x) = x^{\frac{3}{2}} + x^{\frac{-3}{2}} - 4\left(x + \dfrac{1}{x}\right)$ for all permissible real $x$ is $\boxed{-6}$, which corresponds to **Option 4**.