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pgpilot326Best 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.
 8 months ago

pgpilot326Best ResponseYou've already chosen the best response.0
in the second function
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
i did i got it wrong.. :/
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
can you show me how you would do it?
 8 months ago

pgpilot326Best 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?
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
yes, i missed one step, thank you.
 8 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
I got it, thanks alot.
 8 months ago

surjithayerBest 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
 8 months ago
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