009A Sample Final 1, Problem 2 Detailed Solution

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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)}   is differentiable at  $x=3$.

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Background Information:
1.  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f(x)}   is continuous at  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=a}   if
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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 at  $x=3.$

(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
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 \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
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_{h\rightarrow 0^-}\frac{f(3+h)-f(3)}{h}=\lim_{h\rightarrow 0^+}\frac{f(3+h)-f(3)}{h},}
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  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=3.}

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

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