Question

y''=\frac{ x ^{3}-2x ^{2} -x+2}{x-2}

Answer 1

\[ y''=\frac{ x ^{3}-2x ^{2} -x+2}{x-2}
\]

Answer 2

\[y'' = ? , y=\frac{ x ^{3}-2x ^{2} -x+2}{ x-2 }\]

Answer 3

divide and then integrate i guess

Answer 4

oh divide and then differentiate might be easiest

Answer 5

you might note that
\[\frac{ x ^{3}-2x ^{2} -x+2}{ x-2 }=x^2-1\]

Answer 6

then take the derivative twice, get \(2\)

Answer 7

so long as \(x\neq 2\) that is

Answer 8

really , thats explain the x=0 (y')