Find all possible local extrema using the first derivative test

Question

f'(x)=-12+12x-3x²

 

 

Answer 1

When we find the local extremas, we are finding the maximum and minimum points.  This is done by setting the derivative of the function to zero, since the slope of the line tangent to the maximum or minimum is zero.

 

f'(x) = 0

 

12 - 6x = 0

 

6(2 - x) = 0

 

x = 2

 

The x-value is the location of the maximum or minimum.  To find the max or min, we perform a test point using x=1 and x=3 and evaluate them into the first derivative.  If the derivative is positive, it means that the graph is increasing at the interval.  If it is negative, it means that the graph is decreasing at the interval.

 

f'(1) = 6(2 - 1)

       = 6(1)

       = 6

 

f'(3) = 6(2 - 3)

       = 6(-1)

       = -6

 

f'(1) is positive.  The graph increases at the interval (-∞, 2).  f'(3) is negative.  The graph decreases at the interval (2, ∞).  Therefore, we will have a maximum at x=2.

 

Evaluate f(2) to find the value of the maximum.

 

f(2) = -12 + 12(2) - 3(2)²

      = -12 + 24 - 12

      = 0

 

The maximum is (2, 0).