The function g is given by g(x)=4x^3+3x^2-6x+1. what is the absolute minimum value of g on the closed interval [-2,1]?

The smallest value of the output of the given function on the interval [-2, 1]

is the absolute minimum of the function in the interval.

The absolute minimum on the closed interval [-2, 1] is

Reasons:

The given function is g(x) = 4·x³ + 3·x² - 6·x + 1

At the extremum (maximum and minimum) points, we have;

2·x² + x - 1 = 0

x² + 0.5·x - 0.5 = 0

Using a graphing calculator, we get;

(x + 1)·(x - 0.5) = 0

x = -1, and x = 0.5 both values are in the given interval [-2, 1]

At x = -1, g(-1) = 4×(-1)³ + 3×(-1)² - 6×(-1) + 1 = 6

At x = (0.5), g(0.5) = 4×(0.5)³ + 3×(0.5)² - 6×(0.5) + 1 = -0.75

Therefore;

The local minimum on the closed interval [-2, 1] is (0.5, -0.75)

However from plotting the graph, we have the following table of values;

The absolute minimum is therefore, the point

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y=3x+4

step-by-step explanation:

given the points a(0,4) and b(-4, -8)

eqn of line ab: (y-4)/(x-0)=(-8-4)/(-4-0)

(y-4)/x=-12/-4=3

y=3x+4