009A Sample Final 1, Problem 2

Consider the following piecewise defined function:

$f(x)=\left\{{\begin{array}{lr}x+5&{\text{if }}x<3\\4{\sqrt {x+1}}&{\text{if }}x\geq 3\end{array}}\right.$

a) Show that $f(x)$ is continuous at $x=3$.

b) Using the limit definition of the derivative, and computing the limits from both sides, show that $f(x)$ is differentiable at $x=3$.

Foundations:

Solution:

(a)

Step 1:
We first calculate $\lim _{x\rightarrow 3^{+}}f(x)$. We have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 3^{+}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 3}4{\sqrt {x+1}}}\\&&\\&=&\displaystyle {4{\sqrt {3+1}}}\\&&\\&=&\displaystyle {8}\end{array}}$

Step 2:
Now, we calculate $\lim _{x\rightarrow 3^{-}}f(x)$. We have
${\begin{array}{rcl}\displaystyle {\lim _{x\rightarrow 3^{-}}f(x)}&=&\displaystyle {\lim _{x\rightarrow 3}x+5}\\&&\\&=&\displaystyle {3+5}\\&&\\&=&\displaystyle {8}\end{array}}$

Step 3:
Now, we calculate $f(3)$. We have
$f(3)=4{\sqrt {3+1}}=8$.
Since $\lim _{x\rightarrow 3^{+}}f(x)=\lim _{x\rightarrow 3^{-}}f(x)=f(3)$, $f(x)$ is continuous.

(b)

Step 1:

Step 2:

Step 3:

Final Answer:
(a) Since $\lim _{x\rightarrow 3^{+}}f(x)=\lim _{x\rightarrow 3^{-}}f(x)=f(3)$, $f(x)$ is continuous.
(b)

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