complete the definition of the (h(x)) so that it is continuous over its domain.\nh(x)=\begin{cases}x^{3},&x…

complete the definition of the (h(x)) so that it is continuous over its domain.\nh(x)=\begin{cases}x^{3},&x < 0\\a,&x = 0\\sqrt{x},&0 < x < 4\\b,&x = 4\\4-\frac{1}{2}x,&x>4end{cases}\n(a=)\n(b=)

complete the definition of the (h(x)) so that it is continuous over its domain.\nh(x)=\begin{cases}x^{3},&x < 0\\a,&x = 0\\sqrt{x},&0 < x < 4\\b,&x = 4\\4-\frac{1}{2}x,&x>4end{cases}\n(a=)\n(b=)

Answer

Explanation:

Step1: Find the left - hand limit as x approaches 0

For (x<0), (h(x)=x^{3}). (\lim_{x\rightarrow0^{-}}h(x)=\lim_{x\rightarrow0^{-}}x^{3}=0). For the function to be continuous at (x = 0), (h(0)=a) must equal the left - hand limit. So (a = 0).

Step2: Find the left - hand and right - hand limits as x approaches 4

For (0<x<4), (h(x)=\sqrt{x}). (\lim_{x\rightarrow4^{-}}h(x)=\lim_{x\rightarrow4^{-}}\sqrt{x}=\sqrt{4}=2). For (x > 4), (h(x)=4-\frac{1}{2}x). (\lim_{x\rightarrow4^{+}}h(x)=\lim_{x\rightarrow4^{+}}(4-\frac{1}{2}x)=4-\frac{1}{2}\times4=4 - 2=2). For the function to be continuous at (x = 4), (h(4)=b) must equal the left - hand and right - hand limits. So (b = 2).

Answer:

(a = 0) (b = 2)