Find the maximum and minimum of the function f (x)=x^3-2x^2-5x+6 on the interval (-3,3)

Question

Find the maximum and minimum values for the function given

Answer 1

The derivative f'(x) of f(x) gives us the values of x for max and min.
That is, f'(x)=3x^2-4x-5=0.
The roots of this quadratic are x=(4±sqrt(16+60))/6=2.120 or -0.786.
Both roots are in the given range. To find out which is max or min we can find out f(x) in the vicinity of these values.
At x=2, f(x)=-4, and at x=2.5 f(x)=-3.375; f(-1)=8 and f(-0.5)=7.875.
The values of f(-0.786)=8.209 (max) and f(2.120)=-4.061 (min).
f(x)=0 when x=-2, 1 and 3 so it factorises into:
(x+2)(x-1)(x-3).

Answer 2

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