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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)) :}

$$\:\left(\mathrm{1}\right)\:\underset{{x}\rightarrow−\mathrm{2}^{−} } {\mathrm{lim}}\:\left({ax}+{b}\right)=\:\underset{{x}\rightarrow−\mathrm{2}^{+} } {\mathrm{lim}}\left({x}^{\mathrm{2}} +\mathrm{2}{b}−\mathrm{17}\right) \\ $$$$\:\:\:\:\:\:\:−\mathrm{2}{a}+{b}\:=\:\mathrm{4}+\mathrm{2}{b}−\mathrm{17} \\ $$$$\:\:\:\:\:\:\:−\mathrm{2}{a}−{b}\:=−\mathrm{13} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}\:\left({x}^{\mathrm{2}} +\mathrm{2}{b}−\mathrm{17}\right)=\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}\:\left(\mathrm{3}{x}−\mathrm{3}+{a}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{1}+\mathrm{2}{b}−\mathrm{17}\:=\:\mathrm{3}−\mathrm{3}+{a} \\ $$$$\:\:\:\:\:\:\:\:−{a}+\mathrm{2}{b}\:=\:\mathrm{16}\Rightarrow{a}=\mathrm{2}{b}−\mathrm{16} \\ $$$$\Rightarrow−\mathrm{2}\left(\mathrm{2}{b}−\mathrm{16}\right)−{b}=−\mathrm{13} \\ $$$$\Rightarrow−\mathrm{5}{b}\:=\:−\mathrm{45}\:\Rightarrow\begin{cases}{{b}=\mathrm{9}}\\{{a}=\mathrm{2}}\end{cases} \\ $$

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