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Question
question 6
let $f(x)=\
$$\begin{cases}8x - 3 & \\text{if } x \\leq 5 \\\\-5x + b & \\text{if } x > 5\\end{cases}$$
$
if $f(x)$ is a function which is continuous everywhere, then we must have
b =
question help: video
Step1: Find left limit at x=5
$\lim_{x \to 5^-} f(x) = 8(5) - 3$
Step2: Calculate left limit value
$\lim_{x \to 5^-} f(x) = 40 - 3 = 37$
Step3: Find right limit at x=5
$\lim_{x \to 5^+} f(x) = -5(5) + b$
Step4: Set limits equal for continuity
$37 = -25 + b$
Step5: Solve for b
$b = 37 + 25$
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