009A Sample Midterm 3, Problem 2 Detailed Solution

Sketch the graph of  $f.$  At each point of discontinuity, state whether  $f$  is left or right continuous.

$f(x)={\begin{array}{cc}{\Bigg \{}&{\begin{array}{cc}x^{3}+1&x\leq 0\\-x+1&0<x<2\\-x^{2}+10x-15&x\geq 2\end{array}}\end{array}}$

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Background Information:
Suppose  $f(x)$  is discontinuous at  $x=a.$
Now,  $f$  is left continuous at  $x=a$  if  $\lim _{x\rightarrow a^{-}}f(x)=f(a).$
Similarly,  $f$  is right continuous at  $x=a$  if  $\lim _{x\rightarrow a^{+}}f(x)=f(a).$

Solution:

Step 1:
We start by graphing  $f.$
Insert graph

Step 2:
Now, we get a common denominator for the fractions in the numerator.
Hence, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\frac {(1+x+h)3x}{(3x+3h)(3x)}}-{\frac {(1+x)(3x+3h)}{(3x+3h)(3x)}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {\frac {3x+3x^{2}+3xh-(3x+3h+3x^{2}+3hx)}{(3x+3h)(3x)}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {-3h}{h(3x+3h)(3x)}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {-3}{(3x+3h)(3x)}}}\\&&\\&=&\displaystyle {\frac {-3}{(3x)(3x)}}\\&=&\displaystyle {-{\frac {1}{3x^{2}}}.}\end{array}}$

Final Answer:
${\frac {dy}{dx}}=-{\frac {1}{3x^{2}}}$

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