hw7 continuity (target l5; §2.4)\nscore: 6/8 answered: 6/8\nquestion 7\nlet ( f(x)=\begin{cases}8x…
hw7 continuity (target l5; §2.4)\nscore: 6/8 answered: 6/8\nquestion 7\nlet ( f(x)=\begin{cases}8x - 2&\text{if }xleq3\\-7x + b&\text{if }x > 3end{cases})\nif ( f(x) ) is a function which is continuous everywhere, then we must have\n( b=)\nbasic funcs trig\nenter a mathematical expression more_
Answer
Explanation:
Step1: Recall continuity condition
For a function to be continuous at (x = 3), (\lim_{x\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{+}}f(x)=f(3)). First, find (\lim_{x\rightarrow3^{-}}f(x)). When (x\rightarrow3^{-}), (f(x)=8x - 2). So (\lim_{x\rightarrow3^{-}}f(x)=8\times3 - 2). [ \begin{align*} \lim_{x\rightarrow3^{-}}f(x)&=24 - 2\ &=22 \end{align*} ]
Step2: Find (\lim_{x\rightarrow3^{+}}f(x))
When (x\rightarrow3^{+}), (f(x)=-7x + b). So (\lim_{x\rightarrow3^{+}}f(x)=-7\times3 + b=-21 + b).
Step3: Set the two - sided limits equal
Since the function is continuous at (x = 3), we set (\lim_{x\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{+}}f(x)). (22=-21 + b). Solve for (b): [ \begin{align*} b&=22 + 21\ b&=43 \end{align*} ]
Answer:
43