find all critical points

Question

f(x)=10((x-1)^2)e^-x

Answer 1

To find the critical points, we must set the derivative of f(x) equal to zero.  The critical points are the location of maximum and minimum points of a graph.  The max and min have lines whose slope is zero hat are tangential to them.

 

Let's make f(x) easier to find the derivative of.

 

f(x) = 10(x - 1)(x - 1)e^-x

f(x) = 10e^-x(x² - 2x + 1)

 

d/dx[f(x)] = 0

 

-10e^-x (x² - 2x + 1) + 10e^-x (2x - 2) = 0

 

-10e^-x [(x² - 2x + 1) - (2x - 2)] = 0

-10e^-x (x² - 2x + 1 - 2x + 2) = 0

-10e^-x (x² - 4x + 3) = 0

 

Set the term in parenthesis to zero.

 

x² - 4x + 3 = 0

(x - 3)(x - 1) = 0

x = 3   and  x = 1

 

Our critical points are x = 1  and  x = 3.