Question

complete the definition of h(x) so that it is continuous over its domain.
\\h(x)=\

$$\begin{cases} x^3, & x < 0 \\\\ a, & x = 0 \\\\ \\sqrt{x}, & 0 < x < 4 \\\\ b, & x = 4 \\\\ 4 - \\frac{1}{2}x, & x > 4 \\end{cases}$$

\\
\\(a = \square\\)
\\(b = \square\\)

Explanation:

Step1: Find $a$ via left limit at 0

For continuity at $x=0$, $\lim_{x \to 0^-} h(x) = h(0)$.
$\lim_{x \to 0^-} x^3 = 0^3 = 0$, so $a=0$.

Step2: Find $b$ via left limit at 4

For continuity at $x=4$, $\lim_{x \to 4^-} h(x) = h(4)$.
$\lim_{x \to 4^-} \sqrt{x} = \sqrt{4} = 2$, so $b=2$.

Answer:

$a = 0$
$b = 2$