Question

find the absolute maximum and absolute minimum values of the function

$f(x)=x^{3}+12x^{2}-27x + 9$

over each of the indicated intervals.

(a) the interval $=-10,0$.

1. absolute maximum=
2. absolute minimum=

(b) the interval $=-7,2$.

1. absolute maximum=
2. absolute minimum=

(c) the interval $=-10,2$.

1. absolute maximum=
2. absolute minimum=

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=x^{3}+12x^{2}-27x + 9\) is \(f^\prime(x)=3x^{2}+24x-27 = 3(x^{2}+8x - 9)=3(x + 9)(x-1)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \(3(x + 9)(x - 1)=0\). The critical points are \(x=-9\) and \(x = 1\)

Part (a) Interval \([-10,0]\)

Step3: Evaluate the function at critical points and endpoints

• \(f(-10)=(-10)^{3}+12(-10)^{2}-27(-10)+9=-1000 + 1200+270 + 9=479\)
• \(f(-9)=(-9)^{3}+12(-9)^{2}-27(-9)+9=-729+972 + 243+9=504\)
• \(f(0)=0^{3}+12\times0^{2}-27\times0 + 9=9\)

Part (b) Interval \([-7,2]\)

Step4: Evaluate the function at critical points and endpoints

• \(f(-7)=(-7)^{3}+12(-7)^{2}-27(-7)+9=-343+588+189 + 9=443\)
• \(f(1)=1^{3}+12\times1^{2}-27\times1 + 9=1+12-27 + 9=-5\)
• \(f(2)=2^{3}+12\times2^{2}-27\times2 + 9=8+48-54 + 9=11\)

Part (c) Interval \([-10,2]\)

Step5: Evaluate the function at critical points and endpoints

• \(f(-10)=479\) (calculated in part (a))
• \(f(-9)=504\) (calculated in part (a))
• \(f(1)=-5\) (calculated in part (b))
• \(f(2)=11\) (calculated in part (b))

Answer:

(a)

1. Absolute maximum \(=504\)
2. Absolute minimum \(=9\)

(b)

1. Absolute maximum \(=443\)
2. Absolute minimum \(=-5\)

(c)

1. Absolute maximum \(=504\)
2. Absolute minimum \(=-5\)