Question

Perform a first derivative test on the function ​f(x)equals2 x cubed plus 3 x squared minus 120 x plus 6​; ​[minus5​,8​]. Bold

Answer 1

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