For the function f(x)=2x3−3x2−12x+5, find the derivative:
f′(x)=6x2−6x−12
To find critical points, set f′(x)=0:
6x2−6x−12=0
x2−x−2=0
(x−2)(x+1)=0
Critical points: x=2 and x=−1
Use the second derivative test to classify the critical points:
f′′(x)=12x−6
At x=−1:
f′′(−1)=12(−1)−6=−18
Since f′′(−1)<0, x=−1 gives a maximum.
At x=2:
f′′(2)=12(2)−6=18
Since f′′(2)>0, x=2 gives a minimum.
Maximum value at x=−1:
f(−1)=2(−1)3−3(−1)2−12(−1)+5
f(−1)=−2−3+12+5
f(−1)=12
Minimum value at x=2:
f(2)=2(2)3−3(2)2−12(2)+5
f(2)=16−12−24+5
f(2)=−15
The difference between maximum and minimum values:
12−(−15)=12+15=27
Therefore, the difference is 27.