Question

the function g is given by ( g(x)=4x^{3}+3x^{2}-6x + 1 ). what is the absolute minimum value of g on the closed interval ( -2,1 )?

Explanation:

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

The derivative \(g^{\prime}(x)=12x^{2}+6x - 6\). Factor it: \(g^{\prime}(x)=6(2x^{2}+x - 1)=6(2x - 1)(x + 1)\)

Step2: Find the critical points

Set \(g^{\prime}(x)=0\). Then \(2x-1 = 0\) gives \(x=\frac{1}{2}\), and \(x + 1=0\) gives \(x=-1\). Both \(\frac{1}{2}\) and \(-1\) are in the interval \([-2,1]\)

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

• For \(x=-2\): \(g(-2)=4\times(-2)^{3}+3\times(-2)^{2}-6\times(-2)+1=4\times(-8)+3\times4 + 12 + 1=-32+12 + 12 + 1=-7\)
• For \(x=-1\): \(g(-1)=4\times(-1)^{3}+3\times(-1)^{2}-6\times(-1)+1=-4 + 3+6 + 1=6\)
• For \(x=\frac{1}{2}\): \(g(\frac{1}{2})=4\times(\frac{1}{2})^{3}+3\times(\frac{1}{2})^{2}-6\times\frac{1}{2}+1=4\times\frac{1}{8}+3\times\frac{1}{4}-3 + 1=\frac{1}{2}+\frac{3}{4}-2=-\frac{3}{4}\)
• For \(x = 1\): \(g(1)=4\times1^{3}+3\times1^{2}-6\times1+1=4 + 3-6 + 1=2\)

Answer:

A. -7