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 $x=3$ .

1

Foundations:
Recall:
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)} 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=a} if 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 a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).}
2. The definition of derivative 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 f(x)} is 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)=\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.

(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

{\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 $\lim _{h\rightarrow 0^{-}}{\frac {f(3+h)-f(3)}{h}}=\lim _{h\rightarrow 0^{+}}{\frac {f(3+h)-f(3)}{h}},$
$f(x)$ is differentiable at $x=3.$

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