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

1 Answer
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)=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