Q. Let h(x)=2x3+3x2−12x.What is the absolute minimum value of h over the closed interval −3≤x≤3 ?Choose 1 answer:(A) −7(B) −32(C) −45(D) 20
Find Derivative of h(x): To find the absolute minimum value of the function h(x) over the closed interval [−3,3], we need to find the critical points of h(x) within the interval and evaluate h(x) at the endpoints of the interval. Critical points occur where the derivative h′(x) is zero or undefined. Let's find the derivative of h(x).h(x)=2x3+3x2−12xh′(x)=dxd(2x3+3x2−12x)=6x2+6x−12
Find Critical Points: Now we need to find the values of x where h′(x)=0.0=6x2+6x−12Divide by 6 to simplify:0=x2+x−2Factor the quadratic equation:0=(x+2)(x−1)The solutions are x=−2 and x=1.
Evaluate h(x) at Critical Points and Endpoints: We have found two critical points, x=−2 and x=1, within the interval [−3,3]. Now we need to evaluate h(x) at these critical points and at the endpoints of the interval, x=−3 and x=3. Let's calculate h(−3), h(−2), h(1), and x=−20. x=−21
Calculate h(−3): Now let's calculate h(−2):h(−2)=2(−2)3+3(−2)2−12(−2)=2(−8)+3(4)+24=−16+12+24=20
Calculate h(−2): Next, we calculate h(1):h(1)=2(1)3+3(1)2−12(1)=2(1)+3(1)−12=2+3−12=−7
Calculate h(1): Finally, we calculate h(3):h(3)=2(3)3+3(3)2−12(3)=2(27)+3(9)−36=54+27−36=45
Calculate h(3): We have evaluated h(x) at the critical points and the endpoints of the interval. The values are:h(−3)=9h(−2)=20h(1)=−7h(3)=45The absolute minimum value of h(x) over the interval [−3,3] is the smallest of these values, which is −7.