If the function f(x),defined below,is continu | vedclass

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(A) Since $f(x)$ is continuous on $[0, 8]$,it must be continuous at $x = 2$ and $x = 4$.For continuity at $x = 2$:$\lim_{x 	o 2^-} f(x) = \lim_{x 	o

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$a = 3, \ b = -2$

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(A) Since $f(x)$ is continuous on $[0, 8]$,it must be continuous at $x = 2$ and $x = 4$.For continuity at $x = 2$:$\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^+} f(x) = f(2)$$\lim_{x 	o 2^-} (x^2 + ax + b) = 3(2) + 2$$4 + 2a + b = 8$$2a + b = 4 \quad \dots (i)$For continuity at $x = 4$:$\lim_{x 	o 4^-} f(x) = \lim_{x 	o 4^+} f(x) = f(4)$$3(4) + 2 = 2a(4) + 5b$$14 = 8a + 5b \quad \dots (ii)$Multiplying equation $(i)$ by $4$,we get $8a + 4b = 16 \quad \dots (iii)$Subtracting $(iii)$ from $(ii)$:$(8a + 5b) - (8a + 4b) = 14 - 16$$b = -2$Substituting $b = -2$ into $(i)$:$2a - 2 = 4$$2a = 6 \implies a = 3$Thus,$a = 3$ and $b = -2$.

If the function $f(x)$,defined below,is continuous on the interval $[0, 8]$,then
$f(x) = \begin{cases} x^{2} + ax + b, & 0 \le x < 2 \\ 3x + 2, & 2 \le x \le 4 \\ 2ax + 5b, & 4 < x \le 8 \end{cases}$

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If the function $f(x)$,defined below,is continuous on the interval $[0, 8]$,then$f(x) = \begin{cases} x^{2} + ax + b, & 0 \le x < 2 \\ 3x + 2, & 2 \le x \le 4 \\ 2ax + 5b, & 4 < x \le 8 \end{cases}$