Question

given the function ( f(x)=x^{3}+9x^{2}+27x + 7 ), determine the absolute minimum value of ( f ) on the closed interval ( -5,-2 ).

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=x^{3}+9x^{2}+27x + 7\) is \(f^\prime(x)=3x^{2}+18x + 27\).
Factor \(f^\prime(x)\): \(f^\prime(x)=3(x^{2}+6x + 9)=3(x + 3)^{2}\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(3(x + 3)^{2}=0\).
Solving for \(x\), we get \(x=-3\).

Step3: Evaluate the function at the critical point and endpoints

Evaluate \(f(x)\) at \(x=-5\), \(x=-3\), and \(x=-2\).

• For \(x=-5\): \(f(-5)=(-5)^{3}+9(-5)^{2}+27(-5)+7=-125 + 225-135 + 7=-38\).
• For \(x=-3\): \(f(-3)=(-3)^{3}+9(-3)^{2}+27(-3)+7=-27+81 - 81+7=-20\).
• For \(x=-2\): \(f(-2)=(-2)^{3}+9(-2)^{2}+27(-2)+7=-8 + 36-54 + 7=-19\).

Answer:

The absolute minimum value of \(f(x)\) on the interval \([-5,-2]\) is \(-38\).