Find the maximum and minimum values of:

Question

f(x)=x^5e^-x        over [0,10]

Answer 1

To find the maximum and minimum, we take the derivative of f(x) and set it equal to zero.  This is because the slope of the tangent line is the derivative, and the line tangent to the maximum and minimum is zero.  This is known as the first derivative test.

 

d/dx[x⁵e^-x] = 0

 

5x⁴e^-x - x⁵e^-x = 0

 

x⁴e^-x(5 - x) = 0

 

 

x = 0       and       x = 5

 

These x values are the location of the maximum and minimum.  Next, we perform test points using these values.  Since the boundary is between 0 and 10 included, lets choose x=-1 , x=3 ,  x=7, and x=11.  Plug in these values into the derivative. 

 

f'(-1) = (-1)⁴e¹(5 - (-1))

         = 1e¹(6)

         = positive value

 

f'(3) = 3⁴e^-3(5 - 3)

       = 3⁴e^-3(2)

       = positive value

 

f'(7) = 7⁴e^-7(5 - 7)

       = 7⁴e^-7(-2)

       = negative value

 

f'(11) = 11⁴e^-11(5 - 11)

         = 11⁴e^-11(-6)

         = negative value

 

The derivative before x=0 is positive.  The derivative after x=0 is positive.  Therefore, there is no max or min since the values do not change from positive to negative.

 

The derivative before x=5 is positive.  The derivative after x=5 is negative.  This indicates a maximum.

 

There is not max or min at x=10 for the same reasons as x=0.   The value is a constant negative.

 

 

Plug in x=5 into f(x) to get the maximum value.

 

f(5) = 5⁵e^-5

       = 21.06

 

The maximum value is (5, 21.06).