Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 − 9x2 − 216x + 3, [−4, 5]

absolute minimum = -749 and

absolute maximum = 467

Step-by-step explanation:

To get the absolute maximum and minimum of the function, the following steps must be followed.

First, we need to find the values of the function at the given interval [-4, 5].

Given the function f(x) = 6x³ − 9x² − 216x + 3

at x = -4;

f(-4) = 6(-4)³ − 9(-4)² − 216(-4) + 3

f(-4) = 6(-64) - 9(16)+864+3

f(-4) = -256- 144+864+3

f(-4) = 467

at x = 5;

f(5) = 6(5)³ − 9(5)² − 216(5) + 3

f(5) = 6(125) - 9(25)-1080+3

f(5) = 750- 225-1080+3

f(5) = -552

Then we will get the values of the function at the crirical points.

The critical points are the value of x when df/dx = 0

df/dx = 18x²-18x-216 = 0

18x²-18x-216 = 0

Dividing through by 18 will give;

x²-x-12 = 0

On factorizing the resulting quadratic equation;

(x²-4x)+(3x-12) = 0

x(x-4)+3(x-4) = 0

(x+3)(x-4) = 0

x+3 = 0 and x-4 = 0

x = -3 and x = 4 (critical points)

at x  = -3;

f(-3) = 6(-3)³ − 9(-3)² − 216(-3) + 3

f(-3) = 6(-27) - 9(9)+648+3

f(-3) = -162-81+648+3

f(-3) = 408

at x = 4

f(4) = 6(4)³ − 9(4)² − 216(4) + 3

f(4) = 6(64) - 9(16)-864+3

f(4) = 256- 144-864+3

f(4) = -749

Based on the values gotten, it can be seen that the absolute minimum and maximum are -749 and 467 respectively

b

step-by-step explanation:

113.1

step-by-step explanation:

v=4

3πr3=4

3·π·33≈113.09734