Question

Given \( f(x) = \frac{4x + 1}{4} \) and \( g(x) = \sqrt{x^3} \), then \( (g \circ f^{-1}) \left( \frac{3}{8} \right) = \)?

• A. \( \frac{\sqrt{3}}{8} \)
• B. \( \frac{\sqrt{3}}{16} \)
• C. \( \frac{\sqrt{2}}{16} \)
• D. \( \frac{\sqrt{2}}{32} \)

Hint:

For solving inverse function problems, first express the function in terms of \( x \), then solve for \( x \) to find the inverse.

Answer 1

The Correct Option is D

We are given $f(x) = \dfrac{4x + 1}{4}$ and $g(x) = \sqrt{x^3}$, and asked to find $(g \circ f^{-1}) \left( \dfrac{3}{8} \right)$.
 
To begin, we need to find the inverse of $f(x)$, which is $f^{-1}(x)$.
Let $y = f(x) = \dfrac{4x + 1}{4}$, solving for $x$ in terms of $y$: \[ y = \frac{4x + 1}{4} \] \[ 4y = 4x + 1 \] \[ 4x = 4y - 1 \] \[ x = \frac{4y - 1}{4} \] So, $f^{-1}(x) = \dfrac{4x - 1}{4}$.

Now, we apply this inverse to $g(f^{-1}(x))$. Substituting $f^{-1} \left( \dfrac{3}{8} \right)$ into $g(x) = \sqrt{x^3}$: \[ f^{-1} \left( \frac{3}{8} \right) = \frac{4 \cdot \frac{3}{8} - 1}{4} = \frac{\frac{12}{8} - 1}{4} = \frac{\frac{4}{8}}{4} = \frac{1}{8} \] \[ g \left( f^{-1} \left( \frac{3}{8} \right) \right) = \sqrt{\left( \frac{1}{8} \right)^3} = \sqrt{\frac{1}{512}} = \frac{1}{\sqrt{512}} = \frac{\sqrt{2}}{32} \] Thus, the answer is $\dfrac{\sqrt{2}}{32}$.