Question

question given the function ( f(x)=x^{4}+8 x^{3}+18 x^{2}-1 ), determine the absolute minimum value of ( f ) on the closed interval ( -4,1 ).

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=x^{4}+8x^{3}+18x^{2}-1\) is \(f^\prime(x)=4x^{3}+24x^{2}+36x = 4x(x^{2}+6x + 9)=4x(x + 3)^{2}\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \(4x(x + 3)^{2}=0\).
Solving this equation gives \(x = 0\) or \(x=-3\). Both \(x = 0\) and \(x=-3\) are in the interval \([-4,1]\).

Step3: Evaluate the function at critical points and endpoints

• When \(x=-4\), \(f(-4)=(-4)^{4}+8\times(-4)^{3}+18\times(-4)^{2}-1=256-512 + 288-1=31\).
• When \(x=-3\), \(f(-3)=(-3)^{4}+8\times(-3)^{3}+18\times(-3)^{2}-1=81-216+162 - 1=26\).
• When \(x = 0\), \(f(0)=0^{4}+8\times0^{3}+18\times0^{2}-1=-1\).
• When \(x = 1\), \(f(1)=1^{4}+8\times1^{3}+18\times1^{2}-1=1 + 8+18-1=26\).

Answer:

The absolute minimum value of \(f(x)\) on the interval \([-4,1]\) is \(-1\).