# "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)." #