# 009A Sample Final A, Problem 3

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.
Version I:
Version II: