Question

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

Explanation:

Step1: Find the derivative of the function

Using the power rule \((x^n)^\prime=nx^{n - 1}\), for \(y = f(x)=x^{3}+3x^{2}+3x - 1\), the derivative \(f^\prime(x)=3x^{2}+6x + 3\).
Factor out the common factor: \(f^\prime(x)=3(x^{2}+2x + 1)=3(x + 1)^{2}\).

Step2: Find the critical points

Set \(f^\prime(x)=0\), so \(3(x + 1)^{2}=0\).
Solve for \(x\): \(x=-1\) (since \((x + 1)^{2}=0\) implies \(x=-1\)).

Step3: Evaluate the function at the critical point and endpoints

• Evaluate \(f(x)\) at \(x=-3\):

\(f(-3)=(-3)^{3}+3(-3)^{2}+3(-3)-1=-27 + 27-9 - 1=-10\).

• Evaluate \(f(x)\) at \(x=-1\):

\(f(-1)=(-1)^{3}+3(-1)^{2}+3(-1)-1=-1 + 3-3 - 1=-2\).

• Evaluate \(f(x)\) at \(x = 1\):

\(f(1)=1^{3}+3(1)^{2}+3(1)-1=1 + 3+3 - 1=6\).

Answer:

The absolute minimum value of \(f(x)\) on the interval \([-3,1]\) is \(-10\).