Q. Let g(x)=x3−12x+7.The absolute maximum value of g over the closed interval [−4,5] occurs at what x-value?Choose 1 answer:(A) 5(B) −2(C) 2(D) −4
Calculate Derivative: To find the absolute maximum value of the function g(x)=x3−12x+7 on the closed interval [−4,5], we first need to find the critical points of g(x) within the interval. Critical points occur where the derivative g′(x) is zero or undefined.Let's calculate the derivative g′(x).g′(x)=dxd(x3−12x+7)=3x2−12.
Find Critical Points: Now we set the derivative equal to zero to find the critical points.0=3x2−12x2=4x=±2Both x=2 and x=−2 are within the interval [−4,5].
Evaluate Function: Next, we evaluate the function g(x) at the critical points and at the endpoints of the interval to determine the absolute maximum.g(−4)=(−4)3−12(−4)+7=−64+48+7=−9g(−2)=(−2)3−12(−2)+7=−8+24+7=23g(2)=(2)3−12(2)+7=8−24+7=−9g(5)=(5)3−12(5)+7=125−60+7=72
Determine Absolute Maximum: Comparing the values of g(x) at the critical points and endpoints, we find that the absolute maximum value occurs at x=5, which is 72.