Question

given the function ( f(x)=x^{4}+8x^{3}+18x^{2}+10 ), determine the absolute minimum value of ( f ) on the closed interval ( -4,1 ).

Explanation:

Step1: Find the derivative of \(f(x)\)

Use the power rule \((x^n)^\prime=nx^{n - 1}\).
\(f^\prime(x)=4x^{3}+24x^{2}+36x=4x(x^{2}+6x + 9)=4x(x + 3)^{2}\)

Step2: Find the critical points

Set \(f^\prime(x)=0\).
Since \(4x(x + 3)^{2}=0\), then \(x = 0\) or \(x=-3\) (both are in the interval \([-4,1]\))

Step3: Evaluate \(f(x)\) at the critical points and endpoints

• When \(x=-4\):

\(f(-4)=(-4)^{4}+8(-4)^{3}+18(-4)^{2}+10=256-512 + 288+10=42\)

• When \(x=-3\):

\(f(-3)=(-3)^{4}+8(-3)^{3}+18(-3)^{2}+10=81-216+162 + 10=37\)

• When \(x = 0\):

\(f(0)=0^{4}+8\times0^{3}+18\times0^{2}+10=10\)

• When \(x = 1\):

\(f(1)=1^{4}+8\times1^{3}+18\times1^{2}+10=1+8+18+10=37\)

Answer:

The absolute minimum value of \(f(x)\) on the interval \([-4,1]\) is \(10\)