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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is differentiable at $x=3$.

Foundations:
1. $f(x)$  is continuous at $x=a$  if $\lim _{x\rightarrow a^{+}}f(x)=\lim _{x\rightarrow a^{-}}f(x)=f(a).$
2. The definition of derivative for $f(x)$  is  $f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}.$

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:
We need to use the limit definition of derivative and calculate the limit from both sides. So, we have
${\begin{array}{rcl}\displaystyle {\lim _{h\rightarrow 0^{-}}{\frac {f(3+h)-f(3)}{h}}}&=&\displaystyle {\lim _{h\rightarrow 0^{-}}{\frac {(3+h)+5-8}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0^{-}}{\frac {h}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0^{-}}1}\\&&\\&=&\displaystyle {1.}\end{array}}$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {\lim _{h\rightarrow 0^{+}}{\frac {f(3+h)-f(3)}{h}}}&=&\displaystyle {\lim _{h\rightarrow 0^{+}}{\frac {4{\sqrt {3+h+1}}-8}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0^{+}}{\frac {4({\sqrt {4+h}}-{\sqrt {4}})}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0^{+}}{\frac {4({\sqrt {4+h}}-{\sqrt {4}})({\sqrt {4+h}}+{\sqrt {4}})}{h({\sqrt {4+h}}+{\sqrt {4}})}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0^{+}}{\frac {4(4+h-4)}{h({\sqrt {4+h}}+{\sqrt {4}})}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0^{+}}{\frac {4h}{h({\sqrt {4+h}}+{\sqrt {4}})}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0^{+}}{\frac {4}{({\sqrt {4+h}}+{\sqrt {4}})}}}\\&&\\&=&\displaystyle {\frac {4}{2{\sqrt {4}}}}\\&&\\&=&\displaystyle {1.}\\\end{array}}$

Step 3:
Since $\lim _{h\rightarrow 0^{-}}{\frac {f(3+h)-f(3)}{h}}=\lim _{h\rightarrow 0^{+}}{\frac {f(3+h)-f(3)}{h}},$
$f(x)$  is differentiable at $x=3.$

Final Answer:
(a)     Since $\lim _{x\rightarrow 3^{+}}f(x)=\lim _{x\rightarrow 3^{-}}f(x)=f(3),~f(x)$  is continuous.
(b)     Since $\lim _{h\rightarrow 0^{-}}{\frac {f(3+h)-f(3)}{h}}=\lim _{h\rightarrow 0^{+}}{\frac {f(3+h)-f(3)}{h}},$
$f(x)$  is differentiable at $x=3.$

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