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pgpilot326 Group TitleBest ResponseYou've already chosen the best response.0
plug in x = 3. You'll get a number for the first function and set the second function equal to that value. Plug in 3 for x and solve for b.
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.0
in the second function
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
i did i got it wrong.. :/
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
can you show me how you would do it?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@pgpilot326
 11 months ago

pgpilot326 Group TitleBest ResponseYou've already chosen the best response.0
From the first equation:\[4(3)6=126=6\] Thus 7(x)+b = 6 when x=3. So \[7(3)+b=6 \rightarrow 12+b=6 \rightarrow b=21+6=27\] Is this what you got?
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
yes, i missed one step, thank you.
 11 months ago

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
I got it, thanks alot.
 11 months ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
Please note a function is continuos at x=a, if (1) L.H.L=R.H.L (2) \[\lim_{x \rightarrow a} f(x)=f(a)\] \[\lim_{x \rightarrow 3}f \left( x \right)=\lim_{x \rightarrow3}\left( 4x6 \right)\] \[put x=3h,h>0,h \rightarrow 0 as x \rightarrow 3\] \[\lim_{x \rightarrow 3}f \left( x \right)=\lim_{h \rightarrow 0}\left\{ 4\left( 3h \right)6 \right\}=4\left( 30 \right)6=6 \] \[\lim_{x \rightarrow 3+}f \left( x \right)=\lim_{x \rightarrow 3+} \left( 7x+b \right)\] \[put x=3+h,h \rightarrow 0,as x \rightarrow 3+\] \[\lim_{x \rightarrow 3+} f \left( x \right)=\lim_{h \rightarrow 0} \left\{ 7\left( 3+h \right)+b \right\}\]=7(3+0)+b=21+b Because f(x) is continuous at all real values \[Hence \lim_{x \rightarrow3}f(x)=\lim_{x \rightarrow 3+} f(x)\] 6=21+b b=6+21=27
 11 months ago
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