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Question Number 187029 by yaslm last updated on 12/Feb/23

	Answered by cortano12 last updated on 12/Feb/23
	$(1) lim_(x→−2^− ) (ax+b)= lim_(x→−2^+ ) (x^2 +2b−17) −2a+b = 4+2b−17 −2a−b =−13 (2) lim_(x→1^− ) (x^2 +2b−17)=lim_(x→1^+ ) (3x−3+a) 1+2b−17 = 3−3+a −a+2b = 16⇒a=2b−16 ⇒−2(2b−16)−b=−13 ⇒−5b = −45 ⇒ { ((b=9)),((a=2)) :}$
	$(1) \lim_{x \to - 2^{-}} (a x + b) = \lim_{x \to - 2^{+}} (x^{2} + 2 b - 17)$ $- 2 a + b = 4 + 2 b - 17$ $- 2 a - b = - 13$ $(2) \lim_{x \to 1^{-}} (x^{2} + 2 b - 17) = \lim_{x \to 1^{+}} (3 x - 3 + a)$ $1 + 2 b - 17 = 3 - 3 + a$ $- a + 2 b = 16 \Rightarrow a = 2 b - 16$ $\Rightarrow - 2 (2 b - 16) - b = - 13$ $\Rightarrow - 5 b = - 45 \Rightarrow {\begin{cases} b = 9 \\ a = 2 \end{cases}$
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