# 009A Sample Final A, Problem 3

3. (Version I) Consider the following function:  ${\displaystyle f(x)={\begin{cases}{\sqrt {x}},&{\mbox{if }}x\geq 1,\\4x^{2}+C,&{\mbox{if }}x<1.\end{cases}}}$
(a) Find a value of  ${\displaystyle C}$ which makes ${\displaystyle f}$ continuous at ${\displaystyle x=1.}$
(b) With your choice of  ${\displaystyle C}$, is ${\displaystyle f}$ differentiable at ${\displaystyle x=1}$?  Use the definition of the derivative to motivate your answer.

3. (Version II) Consider the following function:  ${\displaystyle g(x)={\begin{cases}{\sqrt {x^{2}+3}},&\quad {\mbox{if }}x\geq 1\\{\frac {1}{4}}x^{2}+C,&\quad {\mbox{if }}x<1.\end{cases}}}$
(a) Find a value of  ${\displaystyle C}$ which makes ${\displaystyle f}$ continuous at ${\displaystyle x=1.}$
(b) With your choice of  ${\displaystyle C}$, is ${\displaystyle f}$ differentiable at ${\displaystyle x=1}$?  Use the definition of the derivative to motivate your answer.

Foundations:
A function ${\displaystyle f}$ is continuous at a point ${\displaystyle x_{0}}$ if
${\displaystyle \lim _{x\rightarrow x_{0}}f(x)=f\left(x_{0}\right).}$
This can be viewed as saying the left and right hand limits exist, and are equal. For problems like these, where we are trying to find a particular value for  ${\displaystyle C}$, we can just set the two descriptions of the function to be equal at the value where the definition of ${\displaystyle f}$ changes.
When we speak of differentiability at such a transition point, being "motivated by the definition of the derivative" really means acknowledge that the derivative is a limit, and for a limit to exist it must agree from the left and the right. This means we must show the derivatives agree for both the descriptions of ${\displaystyle f}$ at the transition point.

Solution:

Version I:
(a) For continuity, we evaluate both rules for the function at the transition point ${\displaystyle x=1}$, set the results equal, and then solve for ${\displaystyle C}$. Since we want
${\displaystyle f(1)\,=\,1\,=\,4\cdot 1^{2}+C,}$
we can set ${\displaystyle C=-3}$, and the function will be continuous (the left and right hand limits agree, and equal the function's value at the point ${\displaystyle x=1}$ ).

(b) To test differentiability, we note that for ${\displaystyle x<1}$,

${\displaystyle f'(x)={\frac {1}{2{\sqrt {x}}}},}$
while for ${\displaystyle x>1}$,
${\displaystyle f'(x)=8x.}$
Thus
${\displaystyle \lim _{x\rightarrow 1^{-}}f'(x)\,=\,{\frac {1}{2{\sqrt {1}}}}\,=\,{\frac {1}{2}},}$
but
${\displaystyle \lim _{x\rightarrow 1^{+}}f'(x)\,=\,8\cdot 1\,=\,8.}$
Since the left and right hand limit do not agree, the derivative does not exist at the point ${\displaystyle x=1}$.
Version II: