Given: x³-2x²-5x+6 on the interval [-3,+3]
the first derivative is: f'(x)=3x²-4x-5
Since f'(x)=0 at extreme points, set f'(x)=0, and find the roots. We have:
x1=(2-√19)/3 (approx.-0.7863) and x2=(2+√19)/3 (approx.2.1196)
So, these two extreme points satisfies the interval: -3<x1,x2<+3
The second derivative is: f''(x)=6x-4 Substitute x1 and x2 into f''(x):
f''(x1)=-2√19<0 and f''(x2)=2√19>0
Since f'(x1)=0 and f''(x1)<0, the curve of f(x) is convex upwards (spills water) around the extreme, and takes the maximum at x=x1. f(x1)=approx.8.2088
Since f'(x2)=0 and f''(x2)>0, the curve of f(x) is concave (holds water) around the extreme, and takes the minimum at x=x2. f(x2)=approx.-4.0607
At the inflection point, f''(x)=0 or x=2/3, the curve changes in form from convex to concave as x increases.
Check the value of f(x) at end points: f(-3)=24 and f(3)=0, so f(-3)<f(x2)<f(3)<f(x1)
Therefore, the answers are:
1. The global maximum is approx.8.2088 at x=(2-√19)/3
2. the local minimum is approx.-4.0607 at x=(2+√19)/3
3. the global minimum is -24 at x=-3