\begin{tabular}{|l|l|l} f′(x)=x3−6x2−15x+20 thru (0,−2) & f′(x)=x​x2+x+1​ thru (0,−3) \\ \hlinef′(x)=4+3x4​ thru (0,2) &

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Begin Tabular L L L F X X3 6x2 15x 20 Thru 0 2 F X X X2 X 1 Thru 0 3 Hlinef X 4 3x4 Thru 0 2 1 (18.76 KiB) Viewed 39 times

\begin{tabular}{|l|l|l} f′(x)=x3−6x2−15x+20 thru (0,−2) & f′(x)=x​x2+x+1​ thru (0,−3) \\ \hlinef′(x)=4+3x4​ thru (0,2) & \\ & f′(x)=2cos(2x) thru (42π​,1) \\ \hline \end{tabular}