Question

given the function $f(x)=x^{3}+9x^{2}+27x + 2$, determine the absolute maximum value of $f$ on the closed interval $-4,-2$.

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=x^{3}+9x^{2}+27x + 2\) 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 \(3(x + 3)^{2}=0\), which gives \(x=-3\).

Step3: Evaluate the function at the critical point and endpoints

• Evaluate \(f(-4)\): \(f(-4)=(-4)^{3}+9(-4)^{2}+27(-4)+2=-64 + 144-108 + 2=-26\).
• Evaluate \(f(-3)\): \(f(-3)=(-3)^{3}+9(-3)^{2}+27(-3)+2=-27+81 - 81+2=-25\).
• Evaluate \(f(-2)\): \(f(-2)=(-2)^{3}+9(-2)^{2}+27(-2)+2=-8 + 36-54 + 2=-24\).

Answer:

The absolute maximum value of \(f(x)\) on the interval \([-4,-2]\) is \(-24\).