009A Sample Final A, Problem 3

Both functions with constants chosen to provide continuity.

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:  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 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 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} 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
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(1)\,=\,1\,=\,4\cdot 1^2+C,}
we can set 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 C=-3} , and the function will be continuous (the left and right hand limits agree, and equal the function's value at the point 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 x=1}  ). (b) To test differentiability, we note that for 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 x>1} ,
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)=\frac{1}{2\sqrt{x}},}
while for 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 x<1} ,
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)=8x.}
Thus
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 \lim_{x\rightarrow 1^+}f'(x)\,=\,\frac{1}{2\sqrt{1}}\,=\,\frac{1}{2},}
but
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 \lim_{x\rightarrow 1^-}f'(x)\,=\,8\cdot1\,=\,8.}
Since the left and right hand limit do not agree, the derivative does not exist at the point 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 x=1} .

Version II:
(a) Like Version I, we begin by setting the two functions equal. We want
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(1)\,=\,\sqrt{1^2+3}\,=\,2\,=\,\frac{\,\,1^2}{4}+C,}
so 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 C=-7/4} makes the function continuous.
(b) We again consider the derivative from each side of 1. For 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 x>1} ,
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)=\frac{1}{2\sqrt{x^2+3}}\cdot 2x\,=\,\frac{x}{\sqrt{x^2+3}},}
while for 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 x<1} ,
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)\,=\,\frac{x}{2}.}
Thus
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 \lim_{x\rightarrow 1^+}f'(x)\,=\,\frac{1}{\sqrt{1^2+3}}\,=\,\frac{1}{2},}
and
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 \lim_{x\rightarrow 1^-}f'(x)\,=\,\frac{1}{2}.}
Since the left and right hand limit do agree, the limit (which is the derivative) does exist at the point 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 x=1} .

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