a. Show that h(x) = x³ / 4 and k(x) = (4x)^(1/3) are inverses of one another.

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Recall that two functions \( h(x) \) and \( k(x) \) are inverses if and only if \( h(k(x)) = x \) and \( k(h(x)) = x \).

Start by computing the composition \( h(k(x)) \). Substitute \( k(x) = (4x)^{1/3} \) into \( h(x) = \frac{x^3}{4} \): \[ h(k(x)) = \frac{\left((4x)^{1/3}\right)^3}{4} \]

Simplify the expression inside \( h(k(x)) \). Since raising to the power 3 and then taking the cube root are inverse operations, simplify \( \left((4x)^{1/3}\right)^3 \) to \( 4x \). Then divide by 4:

\[ h(k(x)) = \frac{4x}{4} \]

Simplify the fraction to get \( h(k(x)) = x \). Next, compute the other composition \( k(h(x)) \) by substituting \( h(x) = \frac{x^3}{4} \) into \( k(x) = (4x)^{1/3} \): \[ k(h(x)) = \left(4 \cdot \frac{x^3}{4}\right)^{1/3} \] Simplify inside the parentheses and then apply the cube root to verify that \( k(h(x)) = x \).

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