## Calculus: Early Transcendentals 8th Edition

$f(-1) = 8$ $f(2)=-19$
Original equation:$f(x)=2x^{3} - 3x^{2}-12x+1$ on the interval $[-2,3]$ To find the absolute maximum and minimum, we first need to find all the critical points of the function, as the local extrema might be the absolute extrema. Critical points of a function occur when a function's derivative is undefined or the derivative is zero. In this case, we see that the function is a polynomial, which means it is defined on all intervals from $(â\infty,\infty)$. Therefore, we need to set the derivative to zero. Finding the derivative of the function through the power rule: $fâ˛(x)=6x^{2}-6x-12$ Then, we set this derivative to zero and solve for x using elementary algebra. We see that the critical point occurs when x=2 $6x^{2}-6x-12=0$ $x^{2}-x-2=0$ $(x+1)(x-2)=0$ $x=-1, 2$ Now we have four possible points of where a absolute minimum or maximum may occur: the critical points $x=-1,2$, as well as the two endpoints of the interval, where $x=-2,3$. To find which one of these is an absolute extrema vs a local extrema, we plug these x values back into the original equation and see what we get. $f(-2)=2(-2)^{3} - 3(-2)^{2}-12(-2)+1=-3$ $f(-1)=2(-1)^{3} - 3(-1)^{2}-12(-1)+1=8$ $f(2)=2(2)^{3} - 3(2)^{2}-12(2)+1=-19$ $f(3)=2(3)^{3} - 3(3)^{2}-12(3)+1=-8$ Of the three values, we see that the minimum y value occurs when x=2 and the maximum y value occurs when x=-1. Final Answer: $f(-1) = 8$ $f(2)=-19$