Complete the definition of the \( h(x) \) so that it is continuous over its domain. \( h(x)=\left\{\begin{array}{ll}x^{3}, & x<0 \\ a, & x=0 \\ \sqrt{x}, & 04\end{array}\right. \) \( a=\square \) \( b=\square \)

To ensure that \( h(x) \) is continuous at \( x = 0 \) and \( x = 4 \), we need the values of \( a \) and \( b \) to match the limits of the functions around these points. 1. At \( x = 0 \): - The limit from the left as \( x \) approaches \( 0 \) is \( \lim_{x \to 0^-} h(x) = 0^3 = 0 \). - The limit from the right as \( x \) approaches \( 0 \) is \( \lim_{x \to 0^+} h(x) = \sqrt{0} = 0 \). - Thus, to ensure continuity at \( x = 0 \), we set \( a = 0 \). 2. At \( x = 4 \): - The limit from the left is \( \lim_{x \to 4^-} h(x) = \sqrt{4} = 2 \). - The limit from the right is \( \lim_{x \to 4^+} h(x) = 4-\frac{1}{2} \times 4 = 2 \). - Hence, to ensure continuity at \( x = 4 \), we set \( b = 2 \). Putting it all together, we have \( a = 0 \) and \( b = 2 \).