## mathcalculus Group Title HELP: CONTINUITY B=? (attached below) 11 months ago 11 months ago

1. mathcalculus Group Title

2. pgpilot326 Group Title

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.

3. pgpilot326 Group Title

in the second function

4. mathcalculus Group Title

i did i got it wrong.. :/

5. mathcalculus Group Title

can you show me how you would do it?

6. mathcalculus Group Title

@pgpilot326

7. pgpilot326 Group Title

From the first equation:$4(3)-6=12-6=6$ 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. mathcalculus Group Title

yes, i missed one step, thank you.

9. mathcalculus Group Title

I got it, thanks alot.

10. surjithayer Group Title

Please note a function is continuos at x=a, if (1) L.H.L=R.H.L (2) $\lim_{x \rightarrow a} f(x)=f(a)$ $\lim_{x \rightarrow 3-}f \left( x \right)=\lim_{x \rightarrow3-}\left( 4x-6 \right)$ $put x=3-h,h>0,h \rightarrow 0 as x \rightarrow 3-$ $\lim_{x \rightarrow 3-}f \left( x \right)=\lim_{h \rightarrow 0}\left\{ 4\left( 3-h \right)-6 \right\}=4\left( 3-0 \right)-6=6$ $\lim_{x \rightarrow 3+}f \left( x \right)=\lim_{x \rightarrow 3+} \left( -7x+b \right)$ $put x=3+h,h \rightarrow 0,as x \rightarrow 3+$ $\lim_{x \rightarrow 3+} f \left( x \right)=\lim_{h \rightarrow 0} \left\{- 7\left( 3+h \right)+b \right\}$=-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