Question

hw7 continuity (target l5; §2.4)
score: 6/8 answered: 6/8
question 7
let ( f(x)=\begin{cases}8x - 2&\text{if }xleq3\\-7x + b&\text{if }x > 3end{cases})
if ( f(x) ) is a function which is continuous everywhere, then we must have
( b=)
basic funcs trig
enter a mathematical expression more_

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = 3\), \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)=f(3)\). First, find \(\lim_{x
ightarrow3^{-}}f(x)\).
When \(x
ightarrow3^{-}\), \(f(x)=8x - 2\). So \(\lim_{x
ightarrow3^{-}}f(x)=8\times3 - 2\).
\[

$$\begin{align*} \lim_{x ightarrow3^{-}}f(x)&=24 - 2\\ &=22 \end{align*}$$

\]

Step2: Find \(\lim_{x

ightarrow3^{+}}f(x)\)
When \(x
ightarrow3^{+}\), \(f(x)=-7x + b\). So \(\lim_{x
ightarrow3^{+}}f(x)=-7\times3 + b=-21 + b\).

Step3: Set the two - sided limits equal

Since the function is continuous at \(x = 3\), we set \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)\).
\(22=-21 + b\).
Solve for \(b\):
\[

$$\begin{align*} b&=22 + 21\\ b&=43 \end{align*}$$

\]

Answer:

43