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