Let
$ f(x) = \begin{cases} 3x & x < 3 \\ a & x = 3 \\ x^2 + b & x > 3 \end{cases} $
Find $a, b$ so that $f(x)$ is continuous at $x=3$, then prove that $f(x)$ is continuous at $x=3$.
Guys, any explanation is helpful.
Let
$ f(x) = \begin{cases} 3x & x < 3 \\ a & x = 3 \\ x^2 + b & x > 3 \end{cases} $
Find $a, b$ so that $f(x)$ is continuous at $x=3$, then prove that $f(x)$ is continuous at $x=3$.
Guys, any explanation is helpful.
$f(x)$ is continuous at $x=3$ if and only if $\lim_{x\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{+}}f(x)=f(3)$, i.e. $\lim_{x\rightarrow3^{-}}3x=\lim_{x\rightarrow3^{+}}x^{2}+b=a$, so $9=9+b=a$, it means that $a=9$ and $b=0$.
The critical point is $x=3$. At this point you want $a = 3x = x^2 + b$ for $f$ to be continuous.
If you draw it, say for $x \in [0,5]$, you will immediately see which value $a = f(3)$ has to be in order for $f$ to be continuous. You really want to draw a picture.
Hope this helps.