let $f(x)=\begin{cases}4x - 5&\text{if }xleq5\\-5x + b&\text{if }x>5end{cases}$. if $f(x)$ is a function…

let $f(x)=\begin{cases}4x - 5&\text{if }xleq5\\-5x + b&\text{if }x>5end{cases}$. if $f(x)$ is a function which is continuous everywhere, then we must have $b = square$. question help: video message instructor submit question jump to answer

let $f(x)=\begin{cases}4x - 5&\text{if }xleq5\\-5x + b&\text{if }x>5end{cases}$. if $f(x)$ is a function which is continuous everywhere, then we must have $b = square$. question help: video message instructor submit question jump to answer

Answer

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = 5$, $\lim_{x\rightarrow5^{-}}f(x)=\lim_{x\rightarrow5^{+}}f(x)$.

Step2: Calculate left - hand limit

$\lim_{x\rightarrow5^{-}}f(x)=4x - 5$. Substitute $x = 5$ into it, we get $4\times5-5=20 - 5=15$.

Step3: Calculate right - hand limit

$\lim_{x\rightarrow5^{+}}f(x)=-5x + b$. Substitute $x = 5$ into it, we get $-5\times5 + b=-25 + b$.

Step4: Set left - hand and right - hand limits equal

Since $\lim_{x\rightarrow5^{-}}f(x)=\lim_{x\rightarrow5^{+}}f(x)$, we have $15=-25 + b$.

Step5: Solve for $b$

Add 25 to both sides of the equation: $b=15 + 25=40$.

Answer:

$40$