Difference between revisions of "009A Sample Final A, Problem 3"

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

3. (Version II) Consider the following function:  $\displaystyle g(x)=\begin{cases} \sqrt{x^{2}+3}, & \quad\mbox{if } x\geq1\\ \frac{1}{4}x^{2}+C, & \quad\mbox{if }x<1. \end{cases}$
(a) Find a value of  $C$ which makes $f$ continuous at $x=1.$ (b) With your choice of  $C$ , is $f$ differentiable at $x=1$ ?  Use the definition of the derivative to motivate your answer.

Foundations:
A function $f$ is continuous at a point $x_{0}$ if
$\lim _{x\rightarrow x_{0}}f(x)=f\left(x_{0}\right).$ 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  $C$ , we can just set the two descriptions of the function to be equal at the value where the definition of $f$ 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 $f$ at the transition point.

Solution:

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

(b) To test differentiability, we note that for $x<1$ ,

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