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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.
 one year ago

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

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

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

mathcalculus Group TitleBest ResponseYou've already chosen the best response.0
@pgpilot326
 one year 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?
 one year ago

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

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