"We will find" \ f'(x), \ "substitute" \ \ x = -1, \ "then look at the sign"
"of" \ \ f'(-1).
\qquad \qquad \qquad \qquad \qquad \qquad f(x) \ = \ { - x^3 + x^2 - x + 7 }/{ x - 2 }.
"Using the Quotient Rule for Derivatives:"
f'(x) \ =
{ ( x - 2 ) [ - x^3 + x^2 - x + 7 ]' - ( - x^3 + x^2 - x + 7 )[ x - 2 ]' }/{ ( x - 2 )^2 }
\qquad \qquad \ \ =
{ ( x - 2 ) [ - 3 x^2 + 2 x - 1 + 0 ] - ( - x^3 + x^2 - x + 7 )\overbrace{ [ 1 - 0 ] }^{1} }/{ ( x - 2 )^2 }
\qquad \qquad \ \ =
\qquad \qquad \qquad \qquad { ( x - 2 ) ( - 3 x^2 + 2 x - 1 ) - ( - x^3 + x^2 - x + 7 ) }/{ ( x - 2 )^2 } .
"No need to multiply out now, and simplify (unless you want"
"to !) -- because we are going to substitute" \ \ x = -1, "and then"
"simplify from there. All we will need to simplify is the"
"numerical expression, not necessarily the polynomial"
"expression. So, we will leave things as they are, and we have:"
f'(x) = \ { ( x - 2 ) ( - 3 x^2 + 2 x - 1 ) - ( - x^3 + x^2 - x + 7 ) }/{ ( x - 2 )^2 } .
"( don't worry about the look below -- it reduces quickly and"
"simply: )"
:. \quad f'(-1) = \ { [ (-1) - 2 ] [ - 3 (-1)^2 + 2 (-1) - 1 ] - [ - (-1)^3 + (-1)^2 - (-1) + 7 ] }/{ ( (-1) - 2 )^2 } .
= { ( -3 ) [ - 3 - 2 - 1 ] - [ - (-1) + 1 + 1 + 7 ] }/{ ( -3 )^2 }
= { ( -3 ) ( -6 ) - ( 10 ) }/{ ( -3 )^2 } \ = \ { 18 - 10 }/{ ( -3 )^2 } \ = \ { "positive" }/{ "positive" } = "positive".
:. \qquad \qquad \qquad \qquad \qquad \qquad f'(-1) \ = \ "positive".
:. \qquad \qquad \qquad \qquad \quad f(x) \ \ "is increasing at" \ \ x = -1.
"This is our answer. "
"Note: we did not need to simplify" \ f'(x), "and afterward, we did"
"not even need to simplify" \ \ f'(-1) \ \ "completely -- all we"
"wanted was the sign of" \ f'(-1), "the actual value of" \ \ f '(-1)
"was not required (though you could compute it, if you"
"wanted)."