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 $=504$

1. absolute minimum $=9$

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

1. absolute maximum $=443$

1. absolute minimum $=-5$

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

1. absolute maximum $=504$

1. absolute minimum $=-5$

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 \(x = 1\)

Step3: Evaluate the function at critical points and endpoints for interval \([-10,2]\)

• For \(x=-10\): \(f(-10)=(-10)^{3}+12(-10)^{2}-27(-10)+9=-1000 + 1200+270 + 9=479\)
• For \(x=-9\): \(f(-9)=(-9)^{3}+12(-9)^{2}-27(-9)+9=-729+972 + 243+9=504\)
• For \(x=1\): \(f(1)=1^{3}+12\times1^{2}-27\times1 + 9=1+12-27 + 9=-5\)
• For \(x = 2\): \(f(2)=2^{3}+12\times2^{2}-27\times2+9=8 + 48-54+9=11\)

Answer:

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