the-following-formulas-for-faf-primeleftarightfa-and-fafprimeleftarightfa-repres

Question

The following formulas for f−′(a)f_{-}^{\prime}\left(aight) and f+′(a)f_{+}^{\prime}\left(aight) represent the left- and right-sided derivatives of a function at a point aa, respectively:f−′(a)=lim⁡h→0−f(a+h)−f(a)hf_{-}^{\prime}\left(aight)={\displaystyle\lim_{h\$$to0^{-}$$}{\frac{f(a+h)-f(a)}{h}}}, f+′(a)=lim⁡h→0+f(a+h)−f(a)hf_{+}^{\prime}\left(aight)={\displaystyle\lim_{h\$$to0^{+}$$}{\frac{f(a+h)-f(a)}{h}}} Consider f(x)={6−x2   if x≤23x−4   if x\u003e2f\left(xight)=\begin{cases}$$6-x^2$$~~~	ext{if}~x\leq{2}\ 3x-4~~~	ext{if}~x\gt{2}\end{cases}. Find f−′(a)f_{-}^{\prime}\left(aight) and f+′(a)f_{+}^{\prime}\left(aight) at a=2a=2.

Accepted Answer

f−′(2)=−4f_{-}^{\prime}\left(2\right)=-4f+′(2)=3f_{+}^{\prime}\left(2\right)=3

Answer

f−′(2)=−4f_{-}^{\prime}\left(2\right)=-4f+′(2)=−3f_{+}^{\prime}\left(2\right)=-3

Answer

f−′(2)=6f_{-}^{\prime}\left(2\right)=6f+′(2)=3f_{+}^{\prime}\left(2\right)=3

Answer

f−′(2)=6f_{-}^{\prime}\left(2\right)=6f+′(2)=−3f_{+}^{\prime}\left(2\right)=-3