find intervals on increasing and decreasing as well as where f(x) is concave up and down

Question

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

 

Please help!

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.

 

To find intervals of increase and decrease, we use test points using x values near the critical points.  Let's use

x=0 , x=2 , and x = 4.  Evaluate them into the derivative. If derivative is positive, the f(x) is increasing.  If derivative is negative, then f(x) is decreasing.

 

f'(0) = -10e^-0 (0² - 4(0) + 3)

       = -10 (3)

       = -30

 

f'(2) = -10e^-2 ((2)² - 4(2) + 3)

      = -10e^-2 (4 - 8 + 3)

      = -10e^-2 (-1)

      = 10e^-2

 

f'(4) = -10e^-4 ((4)² - 4(4) + 3)

      = -10e^-4 (16 - 16 + 3)

      = -10e^-4 (3)

      = -30e^-4

 

f(x) is decreasing at (-∞, 1)

f(x) is increasing at (1, 3)

f(x) is decreasing at (3, ∞)

 

 

Next is to find where f(x) is concave up and concave down.  We take the second derivative of f(x) and set it equal to zero.  When solve for x, we are finding the location of the points of inflection.  A point of inflection is where f(x) changes shape. Once the points of inflection has been found, use values near those points and evaluate the second derivative using those x values.  If the second derivative is positive, then f(x) is concave up.  If second derivative is negative, then f(x) is concave down.

 

I will let you complete these step on your own.  To help you start, take the derivative of this equal to zero:

 

 

d/dx[-10e^-x (x² - 4x + 3)] = 0