If the function f(x) = begin{cases} 5x - 4, & | vedclass

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(A) For the function $f(x)$ to be continuous at every point of its domain,it must be continuous at the point $x = 1$,where the definition of the funct

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$-1$

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(A) For the function $f(x)$ to be continuous at every point of its domain,it must be continuous at the point $x = 1$,where the definition of the function changes. For continuity at $x = 1$,the left-hand limit must equal the right-hand limit and the value of the function at $x = 1$: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$ Calculating the left-hand limit and $f(1)$: $\lim_{x 	o 1^-} (5x - 4) = 5(1) - 4 = 1$ Calculating the right-hand limit: $\lim_{x 	o 1^+} (4x^2 + 3bx) = 4(1)^2 + 3b(1) = 4 + 3b$ Equating the two: $1 = 4 + 3b$ $3b = 1 - 4$ $3b = -3$ $b = -1$ Thus,the value of $b$ is $-1$.

If the function $f(x) = \begin{cases} 5x - 4, & 0 < x \le 1 \\ 4x^2 + 3bx, & 1 < x < 2 \end{cases}$ is continuous at every point of its domain,then the value of $b$ is

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