If the function f(x) is continuous on its dom | vedclass

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(D) For the function to be continuous at $x = 0$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o 0^-} f(x) = f(0)$.$\lim_{x 	o 0^-}

Vedclass · Accepted Answer

$12$

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(D) For the function to be continuous at $x = 0$,the left-hand limit must equal the right-hand limit: $\lim_{x 	o 0^-} f(x) = f(0)$.$\lim_{x 	o 0^-} (\frac{\sin ax}{x} + 3) = 0 + 5$.Since $\lim_{x 	o 0} \frac{\sin ax}{x} = a$,we have $a + 3 = 5$,which gives $a = 2$.For the function to be continuous at $x = 1$,the left-hand limit must equal the right-hand limit: $f(1) = \lim_{x 	o 1^+} f(x)$.$1 + 5 = \sqrt{1^2 + 8} - b$.$6 = \sqrt{9} - b$.$6 = 3 - b$,which gives $b = -3$.Finally,calculating $7a + b + 1 = 7(2) + (-3) + 1 = 14 - 3 + 1 = 12$.

If the function $f(x)$ is continuous on its domain $[-2, 2]$,where $f(x) = \begin{cases} \frac{\sin ax}{x} + 3, & -2 \leq x < 0 \\ x + 5, & 0 \leq x \leq 1 \\ \sqrt{x^2 + 8} - b, & 1 < x \leq 2 \end{cases}$,then $7a + b + 1$ is equal to:

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