find the absolute maximum and minimum values of each function over the indicated interval, and indicate the…

find the absolute maximum and minimum values of each function over the indicated interval, and indicate the x - values at which they occur. f(x)=2x^3 - x^2 - 4x + 6, -1,0 the absolute maximum value is 206/27 at x=-2/3 (use a comma to separate answers as needed. type an integer or a fraction.) the absolute minimum value is at x= (use a comma to separate answers as needed. type an integer or a fraction.)

find the absolute maximum and minimum values of each function over the indicated interval, and indicate the x - values at which they occur. f(x)=2x^3 - x^2 - 4x + 6, -1,0 the absolute maximum value is 206/27 at x=-2/3 (use a comma to separate answers as needed. type an integer or a fraction.) the absolute minimum value is at x= (use a comma to separate answers as needed. type an integer or a fraction.)

Answer

Answer:

The absolute minimum value is $1$ at $x = 0$.

Explanation:

Step1: Find the derivative

$f'(x)=6x^{2}-2x - 4$.

Step2: Set the derivative equal to 0

$6x^{2}-2x - 4 = 0$. Factor out 2: $2(3x^{2}-x - 2)=0$. Then factor the quadratic: $2(3x + 2)(x - 1)=0$. Solving gives $x=-\frac{2}{3},1$. But $x = 1$ is outside the interval $[-1,0]$, so we discard it.

Step3: Evaluate the function at critical and end - points

Evaluate $f(x)=2x^{3}-x^{2}-4x + 6$ at $x=-1,x =-\frac{2}{3},x = 0$. $f(-1)=2(-1)^{3}-(-1)^{2}-4(-1)+6=-2 - 1+4 + 6=7$. $f(-\frac{2}{3})=2(-\frac{2}{3})^{3}-(-\frac{2}{3})^{2}-4(-\frac{2}{3})+6=2(-\frac{8}{27})-\frac{4}{9}+\frac{8}{3}+6=-\frac{16}{27}-\frac{12}{27}+\frac{72}{27}+6=\frac{-16 - 12+72}{27}+6=\frac{44}{27}+6=\frac{44 + 162}{27}=\frac{206}{27}$. $f(0)=2(0)^{3}-(0)^{2}-4(0)+6 = 6$. Comparing these values, the absolute minimum value of $f(x)$ on $[-1,0]$ is $1$ at $x = 0$.