Find the absolute maximum and minimum values over the indicated interval

Question

F(x)=1/3x^3+x^2-8x+2;   (-1,3)

Answer 1

Matthew has the right procedure.  He set the derivative equal to zero because the maximum and minimum are tangent to lines that have a slope of zero.  The slope of the tangent line is the derivative.  I would like to show you the calculations to how he obtained his max and min.

 

x² + 2x - 8 = 0

 

(x + 4)(x - 2) = 0

 

x = -4      and          x = 2

 

So far so good.  These x values are the location of the max and min.  He chose  x=2 to satisfy the boundaries.  Now we have to see if there is max or min at x=2.

 

Next, we need to pick points of interested and plug them into the derivative.  Lets choose x=1 and  x=2.5

 

These x values are less than 2 and greater than 2, but they remain in boundary.

 

f'(1) = (1 + 4)(1 - 2)

       = (5)(-1)

       = -5

 

f'(2.5) = (2.5 + 4)(2.5 - 2)

          = (6.5)(0.5)

          = 3.25

 

From the interval (-1, 2), the derivative is negative. 

From the interval (2, 3), the derivative is positive. 

 

This indicates a minimum at x=2.  But this is local minimum.  To find the global max/min, we need to plug the boundary points and critical point into the original function.

 

f(-1) = (1/3)(-1)³ + (-1)² - 8(-1) + 2

        = (-1/3) + 1 + 8 + 2

        = -(1/3) + 11

        = (-1 + 33) / 3      

        = 32/3

        = 10.66

 

f(2) = (1/3)(2)³ + 2² - 8(2) + 2

      =  (8/3) + 4 - 16 + 2

      = (8/3) - 10

      = (8 - 30) / 3

      = -22/3

      =  -7.33

 

f(3) = (1/3)(3)³ + 3² - 8(3) + 2

       = 9 + 9 - 24 + 2

       = 18 - 22

       = -4

 

Since x=-1 is not included in the interval, the global maximum is near the point (-1, 10.66) rather than on the point.  The lowest f(x) value is -7.33, so the global minimum is (2, -7.33)

 

 

Answer 2

f'(x)= x²+ 2x-8

 

so if we set f'(x)= 0 = x²+2x-8  the critical points are when x= -4, and 2

so since 2 is between (-1,3) and 2 most be a max or a minimum, after 2 the y values are increasing therefore, it must be the absolute minimum and the absolute max will be at -1.