# Is f(x)=-x/e^(x^2-3x+2)  increasing or decreasing at x=4 ?

Jun 11, 2017

$f$ is increasing at $x = 4$

#### Explanation:

Since the derivative of a function $f$, denoted $f '$, is the "instantaneous rate of change" of the function. A way to tell the behavior of a function around a point is to take the derivative. If $f ' > 0$ then $f$ is increasing. If $f ' < 0$ then $f$ is decreasing.

Since ${x}^{- r} = \frac{1}{x} ^ r$, we can say

$f \left(x\right) = - \frac{x}{{e}^{{x}^{2} - 3 x + 2}} = - x \left({e}^{- {x}^{2} + 3 x - 2}\right)$

And since the Product rule states

$\left[h \left(x\right) g \left(x\right)\right] ' = h ' \left(x\right) g \left(x\right) + h \left(x\right) g ' \left(x\right)$

we can say

$f ' \left(x\right) = - {e}^{- {x}^{2} + 3 x - 2} - x \left({e}^{- {x}^{2} + 3 x - 2}\right) '$

and since the chain rule states $\left(h \left(g \left(x\right)\right)\right) ' = h ' \left(g \left(x\right)\right) g ' \left(x\right)$

we can say

$\left({e}^{- {x}^{2} + 3 x - 2}\right) ' = \left(- {x}^{2} + 3 x - 2\right) ' {e}^{- {x}^{2} + 3 x - 2}$

$= - 2 x + 3 \left({e}^{- {x}^{2} + 3 x - 2}\right) = - 2 x {e}^{- {x}^{2} + 3 x - 2} + 3 {e}^{- {x}^{2} + 3 x - 2}$

Then

$f ' \left(x\right) = - {e}^{- {x}^{2} + 3 x - 2} + 2 {x}^{2} {e}^{- {x}^{2} + 3 x - 2} - 3 x {e}^{- {x}^{2} + 3 x - 2}$

Now to check for the point $x = 4$ plug in $4$

Then

$f ' \left(4\right) = - {e}^{- 16 + 12 - 2} + 32 {e}^{- 16 + 12 - 2} - 12 {e}^{- 16 + 12 - 2}$

$= - {e}^{-} 6 + 32 {e}^{-} 6 - 12 {e}^{-} 6 = 19 {e}^{-} 6$

and since $f ' \left(4\right) = 19 {e}^{-} 6$ then $f ' \left(4\right) > 0$ $\therefore$ $f$ is increasing at the point $x = 4$