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Question Number 105383 by bemath last updated on 28/Jul/20

Given  { ((lim_(x→5) ((f(x)−a)/(x−5)) = 8)),((lim_(x→5) ((x^2 −ax+b)/(f(x)−a)) = 1)) :}  find the value of b+23

$$\mathcal{G}{iven}\:\begin{cases}{\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\frac{{f}\left({x}\right)−{a}}{{x}−\mathrm{5}}\:=\:\mathrm{8}}\\{\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} −{ax}+{b}}{{f}\left({x}\right)−{a}}\:=\:\mathrm{1}}\end{cases} \\ $$$${find}\:{the}\:{value}\:{of}\:{b}+\mathrm{23}\: \\ $$

Answered by john santu last updated on 28/Jul/20

(1) lim_(x→5) ((f(x)−a)/(x−5))=8 → { ((f(5)=a)),((f ′(5)=8)) :}  (2) lim_(x→5) ((x^2 −ax+b)/(f(x)−a)) = 1   ⇒25−5f(5)+b = 0; b = 5f(5)−25  ⇒lim_(x→5) ((2x−a)/(f ′(x))) = 1 ; 10−a=f ′(5)  a = 10−f ′(5)=10−8=2  and b = 5f(5)−25 =  5×2−25=−15  we conclude b+23 =  −15+23=8

$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\frac{{f}\left({x}\right)−{a}}{{x}−\mathrm{5}}=\mathrm{8}\:\rightarrow\begin{cases}{{f}\left(\mathrm{5}\right)={a}}\\{{f}\:'\left(\mathrm{5}\right)=\mathrm{8}}\end{cases} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} −{ax}+{b}}{{f}\left({x}\right)−{a}}\:=\:\mathrm{1}\: \\ $$$$\Rightarrow\mathrm{25}−\mathrm{5}{f}\left(\mathrm{5}\right)+{b}\:=\:\mathrm{0};\:{b}\:=\:\mathrm{5}{f}\left(\mathrm{5}\right)−\mathrm{25} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\frac{\mathrm{2}{x}−{a}}{{f}\:'\left({x}\right)}\:=\:\mathrm{1}\:;\:\mathrm{10}−{a}={f}\:'\left(\mathrm{5}\right) \\ $$$${a}\:=\:\mathrm{10}−{f}\:'\left(\mathrm{5}\right)=\mathrm{10}−\mathrm{8}=\mathrm{2} \\ $$$${and}\:{b}\:=\:\mathrm{5}{f}\left(\mathrm{5}\right)−\mathrm{25}\:= \\ $$$$\mathrm{5}×\mathrm{2}−\mathrm{25}=−\mathrm{15} \\ $$$${we}\:{conclude}\:{b}+\mathrm{23}\:= \\ $$$$−\mathrm{15}+\mathrm{23}=\mathrm{8}\: \\ $$

Commented by bemath last updated on 28/Jul/20

jooss

$${jooss} \\ $$

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