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 [tex]\underline{(-2, -7)}[/tex]
Reasons:
The given function is g(x) = 4·x³ + 3·x² - 6·x + 1
At the extremum (maximum and minimum) points, we have;
[tex]\dfrac{d\left(g(x) \right)}{dx} = \dfrac{d\left(4 \cdot x^3 + 3 \cdot x^2 - 6 \cdot x + 1\right)}{dx} = 12 \cdot x^2 + 6 \dot x - 6 = 0[/tex]
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;
[tex]\begin{array}{|cc|c|}x&&g(x)\\-2&&-7\\-1&&6\\0&&1\\1&&2\end{array}\right][/tex]
The absolute minimum is therefore, the point [tex]\underline{(-2, -7)}[/tex]
Learn more here:
https://brainly.com/question/9581369
https://brainly.com/question/24260561
