Question

precal question: please help!

Answer 1

I think there is one discontinuity

Answer 2

\[1.find~ the~ limit~ at` x \rightarrow-3-~ and ~x \rightarrow-3+,and f(-3)\]
\[2.find~\lim_{x \rightarrow 3-}~and ~\lim_{x \rightarrow 3+},also~f(3)\]
\[3.find~\lim_{x \rightarrow 5-}~and~\lim_{x \rightarrow 5+},also~f(5)\]

Answer 3

how do you find the limit?

Answer 4

\[\lim_{x \rightarrow -3-}f(x)=\lim_{x \rightarrow -3-}\left\{ 3e ^{x+3}+3 \right\}\]
put x=-3-h,h>0,\[h \rightarrow0 ~as~x \rightarrow-3-\]
\[\lim_{x \rightarrow -3-}f(x)=\lim_{h \rightarrow 0}\left\{ 3e ^{-3-h+3}+3 \right\}=\lim_{h \rightarrow 0}\left\{ 3e ^{-h}+3 \right\}\]
\[=\lim_{h \rightarrow 0}\left( \frac{ 3 }{ e^h }+3 \right)=\frac{ 3 }{ e^0 }+3=3+3=6\]
similarly find other limits

Answer 5

\[\lim_{x \rightarrow -3+}f(x)=\lim_{x \rightarrow -3+}\left\{ \frac{ 2 }{ 3 }x^2 \right\}=\frac{ 2 }{ 3 }\left( -3\right)^2=\frac{ 2 }{ 3 }*(-3)^2=6\]
\[f(-3)=\frac{ 2 }{ 3 }(-3)^2=6\]
so \[\lim_{x \rightarrow -3-}f(x)=\lim_{x \rightarrow -3+}f(x)=f(-3)=6\]
hence f(x) is continuous at x=-3

Answer 6

similarly see other points.