Answer:
a) Critical points
x = 4 and x = -5
b) x = 4 corresponds to a minimum point for the function f(x)
x = - 5 corresponds to a maximum point for the function f(x)
c) The minimum value of f(x) in the interval = -298
The maximum value of f(x) in the interval = 431;
Step-by-step explanation:
f(x) = 2x³ + 3x² - 120x + 6 in the interval [-5, 8]
a) To obtain the critical points, we need to obtain the first derivative of the function with respect to x. Because at critical points of a function, f(x), (df/dx) = 0
f'(x) = (df/dx) = 6x² + 6x - 120 = 0
6x² + 6x - 120 = 0
Solving the quadratic equation,
x = 4 or x = -5
The two critical points, x = 4 and x = -5 are in the interval given [-5, 8] (the interval includes -5 and 8, because it's a closed interval)
b) To investigate the nature of the critical points, we obtain f"(x)
Because f'(x) changes sign at the critical points
If f"(x) > 0, then it's a minimum point and if f"(x) < 0 at c, then it's a maximum point.
f"(x) = (d²f/dx²) = 12x + 6
at critical point x = 4
f"(x) = 12x + 6 = 12(4) + 6 = 54 > 0, hence, x = 4 corresponds to a minimum point.
at critical point x = -5
f"(x) = 12x + 4 = 12(-5) + 4 = -56 < 0, hence, x = -5 corresponds to a maximum point.
c) At x = 4,
f(x) = 2x³ + 3x² - 120x + 6 = 2(4)³ + 3(4)² - 120(4) + 6 = -298
At x = - 5
f(x) = 2x³ + 3x² - 120x + 6 = 2(-5)³ + 3(-5)² - 120(-5) + 6 = 431
, you have to take the third root of both sides:![\sqrt[3]{x^{3}} = \sqrt[3]{1}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B1%7D%20)
as a product of 2 polynomials:
is the solution. Another solutions (complex roots) are the roots of quadratic equation.