# Is f(x)=(-x^3+x^2-x+7)/(x-2) increasing or decreasing at x=-1?

Mar 2, 2018

$\setminus q \quad \setminus q \quad \setminus \quad f \left(x\right) \setminus \setminus \setminus \text{is increasing at} \setminus \setminus \setminus x = - 1.$

#### Explanation:

$\text{We will find" \ f'(x), \ "substitute" \ \ x = -1, \ "then look at the sign}$
$\text{of} \setminus \setminus f ' \left(- 1\right) .$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f \left(x\right) \setminus = \setminus \frac{- {x}^{3} + {x}^{2} - x + 7}{x - 2} .$

$\text{Using the Quotient Rule for Derivatives:}$

$f ' \left(x\right) \setminus =$

$\frac{\left(x - 2\right) \left[- {x}^{3} + {x}^{2} - x + 7\right] ' - \left(- {x}^{3} + {x}^{2} - x + 7\right) \left[x - 2\right] '}{{\left(x - 2\right)}^{2}}$

$\setminus q \quad \setminus q \quad \setminus \setminus =$

$\frac{\left(x - 2\right) \left[- 3 {x}^{2} + 2 x - 1 + 0\right] - \left(- {x}^{3} + {x}^{2} - x + 7\right) \setminus {\overbrace{\left[1 - 0\right]}}^{1}}{{\left(x - 2\right)}^{2}}$

$\setminus q \quad \setminus q \quad \setminus \setminus =$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \frac{\left(x - 2\right) \left(- 3 {x}^{2} + 2 x - 1\right) - \left(- {x}^{3} + {x}^{2} - x + 7\right)}{{\left(x - 2\right)}^{2}} .$

$\text{No need to multiply out now, and simplify (unless you want}$
$\text{to !) -- because we are going to substitute" \ \ x = -1, "and then}$
$\text{simplify from there. All we will need to simplify is the}$
$\text{numerical expression, not necessarily the polynomial}$
$\text{expression. So, we will leave things as they are, and we have:}$

$f ' \left(x\right) = \setminus \frac{\left(x - 2\right) \left(- 3 {x}^{2} + 2 x - 1\right) - \left(- {x}^{3} + {x}^{2} - x + 7\right)}{{\left(x - 2\right)}^{2}} .$

$\text{( don't worry about the look below -- it reduces quickly and}$
$\text{simply: )}$

$\therefore \setminus \quad f ' \left(- 1\right) = \setminus \frac{\left[\left(- 1\right) - 2\right] \left[- 3 {\left(- 1\right)}^{2} + 2 \left(- 1\right) - 1\right] - \left[- {\left(- 1\right)}^{3} + {\left(- 1\right)}^{2} - \left(- 1\right) + 7\right]}{{\left(\left(- 1\right) - 2\right)}^{2}} .$

$= \frac{\left(- 3\right) \left[- 3 - 2 - 1\right] - \left[- \left(- 1\right) + 1 + 1 + 7\right]}{{\left(- 3\right)}^{2}}$

 = { ( -3 ) ( -6 ) - ( 10 ) }/{ ( -3 )^2 } \ = \ { 18 - 10 }/{ ( -3 )^2 } \ = \ { "positive" }/{ "positive" } = "positive".

$\therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad f ' \left(- 1\right) \setminus = \setminus \text{positive} .$

$\therefore \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad f \left(x\right) \setminus \setminus \text{is increasing at} \setminus \setminus x = - 1.$

$\text{This is our answer. }$

$\text{Note: we did not need to simplify" \ f'(x), "and afterward, we did}$
$\text{not even need to simplify" \ \ f'(-1) \ \ "completely -- all we}$
$\text{wanted was the sign of" \ f'(-1), "the actual value of} \setminus \setminus f ' \left(- 1\right)$
$\text{was not required (though you could compute it, if you}$
$\text{wanted).}$