Find absolute max and min

Question

Find the absolute maximum and absolute minimum values of f(x)=x^3-3x^2-45x+1 on [-5,3],along with where they occur.

1)the absolute min value is ____ , occurring when x=___

2)the absolute max value is ____, occurring when x=___

Answer 1

For a continuously differentiable function on a closed interval, the absolute max and min or at endpoints or where the first derivative equals 0.

f(x)=x^3-3x^2-45x+1 on [-5,3] meets these conditions.

f'(x) = 3x^2 - 6x -45 = 3(x^2 - 2x - 15) = 3(x-5)(x+3)

This has 0s at 5 and -3. -3 is in the interval, but 5 is not.

Incidentally, it is clear that the max will be at -3, as the function increases to the left and decreases to the right.

f(-5) = (-5)^3 - 3(-5)^2 - 45(-5) + 1 = 26

f(-3) = (-3)^3 - 3(-3)^2 - 45(-3) + 1 = 82

f(3) = (3)^3 - 3(3)^2 - 45(3) + 1 = -134

the global minimum is - 134 at 3.

The global maximum is 82 at x = -3

Answer 2

Find the critical values x= -3,5 and f(xc)= 79, 76

Find f(x) at the end points f(3)= -134 and f(-5)=26

Abs min = (3,134) and abs max=(-3,79)