Question

Find the absolute maximum and absolute minimum values of the function f(x)=x3+6x2−63x+9 over each of the indicated intervals. (a) Interval = [−8,0]. 1. Absolute maximum = -99 2. Absolute minimum = 9 (b) Interval = [−5,4]. 1. Absolute maximum = 348 2. Absolute minimum = -100 (c) Interval = [−8,4]. 1. Absolute maximum = 400 2. Absolute minimum =

Answer 1

Answer:

a) The absolute maximum is 401 and the absolute minimum is 9.

b) The absolute maximum is 349 and the absolute minimum is -99.

c) The absolute maximum is 401 and the absolute minimum is -99.

Step-by-step explanation:

The absolute minimum and absolute maximum values are determined with the help of the First and Second Derivative Tests:

FDT

$3\cdot x^{2} + 12\cdot x - 63 = 0$

The roots of the function are: $x_{1} = 3$ and $x_{2} = -7$. Each point is evaluated in the second derivative of the function:

SDT

$f''(x) = 6\cdot x + 12$

$f''(x_{1}) = 30$ (Absolute minimum)

$f''(x_{2}) = -30$ (Absolute maximum)

The values for each extreme are, respectively:

$f(x_{1}) = -99$

$f(x_{2}) = 401$

Now, each interval is analyzed herein:

a) The absolute maximum is 401 and the absolute minimum is 9.

b) The absolute maximum is 349 and the absolute minimum is -99.

c) The absolute maximum is 401 and the absolute minimum is -99.