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

Question

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

Paisley Johnson · Accepted Answer

#f# is increasing at #x=4#

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)=1/x^r#, we can say

#f(x)=-x/(e^(x^2-3x+2))=-x(e^(-x^2+3x-2))#

And since the Product rule states

#[h(x)g(x)]'=h'(x)g(x)+h(x)g'(x)#

we can say

#f'(x)=-e^(-x^2+3x-2)-x(e^(-x^2+3x-2))'#

and since the chain rule states #(h(g(x)))'=h'(g(x))g'(x)#

we can say

#(e^(-x^2+3x-2))'=(-x^2+3x-2)'e^(-x^2+3x-2)#

#=-2x+3(e^(-x^2+3x-2))=-2xe^(-x^2+3x-2)+3e^(-x^2+3x-2)#

Then

#f'(x)=-e^(-x^2+3x-2)+2x^2e^(-x^2+3x-2)-3xe^(-x^2+3x-2)#

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

Then

#f'(4)=-e^(-16+12-2)+32e^(-16+12-2)-12e^(-16+12-2)#

#=-e^-6+32e^-6-12e^-6=19e^-6#

and since #f'(4)=19e^-6# then #f'(4)>0# #therefore# #f# is increasing at the point #x=4#

Christopher Allers · Answer

undefined To determine whether $ f(x) = -\frac{x}{e^{x^2 - 3x + 2}} $ is increasing or decreasing at $ x = 4 $, we need to examine the sign of its derivative at that point. We'll find the derivative of $ f(x) $ with respect to $ x $ using the quotient rule, then evaluate it at $ x = 4 $ and determine its sign.

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

Now, substituting $ x = 4 $ into $ f'(x) $ gives us:

$$ f'(4) = \frac{-e^{4^2 - 3 \cdot 4 + 2} - 4 \cdot e^{4^2 - 3 \cdot 4 + 2} \cdot (2 \cdot 4 - 3)}{(e^{4^2 - 3 \cdot 4 + 2})^2} $$

$$ f'(4) = \frac{-e^{16 - 12 + 2} - 4 \cdot e^{16 - 12 + 2} \cdot (8 - 3)}{(e^{16 - 12 + 2})^2} $$

$$ f'(4) = \frac{-e^6 - 4 \cdot e^6 \cdot 5}{(e^6)^2} $$

$$ f'(4) = \frac{-e^6 - 20e^6}{e^{12}} $$

$$ f'(4) = \frac{-21e^6}{e^{12}} $$

Since $ e^6 $ and $ e^{12} $ are both positive, the sign of $ f'(4) $ depends only on the sign of $ -21 $. Since $ -21 $ is negative, $ f'(4) $ is negative.

Therefore, $ f(x) $ is decreasing at $ x = 4 $.