009A Sample Final 1, Problem 2

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  $f(x)$  is continuous at  $x=3.$

(b) Using the limit definition of the derivative, and computing the limits from both sides, show that  $f(x)$  is differentiable at  $x=3$.

Foundations:
1.  $f(x)$  is continuous at  $x=a$  if
$\lim _{x\rightarrow a^{+}}f(x)=\lim _{x\rightarrow a^{-}}f(x)=f(a).$
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
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),}
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 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)

Step 1:
We need to use the limit definition of derivative and calculate the limit from both sides. So, 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{(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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