Question

complete the definition of the 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 =
b =

Explanation:

Step1: Find left - hand limit at \(x = 0\)

\(\lim_{x
ightarrow0^{-}}h(x)=\lim_{x
ightarrow0^{-}}x^{3}=0\)
For the function to be continuous at \(x = 0\), \(\lim_{x
ightarrow0^{-}}h(x)=h(0)\), so \(a = 0\).

Step2: Find left - hand limit at \(x = 4\)

\(\lim_{x
ightarrow4^{-}}h(x)=\lim_{x
ightarrow4^{-}}\sqrt{x}=\sqrt{4}=2\)

Step3: Find right - hand limit at \(x = 4\)

\(\lim_{x
ightarrow4^{+}}h(x)=\lim_{x
ightarrow4^{+}}(4-\frac{1}{2}x)=4-\frac{1}{2}\times4=4 - 2=2\)
For the function to be continuous at \(x = 4\), \(\lim_{x
ightarrow4^{-}}h(x)=h(4)\), so \(b = 2\)

Answer:

\(a = 0\), \(b = 2\)