Determine the local extreme values of the function f(x)=x^3-3x.
To find the extreme values first we need to determine f'(x)'s zeros.
x^2= 1 ==> x=1, -1
Thee critical values are 1 and -1
Then extreme values are f(1) and f(-1)
f(1)= 1-3= -2
f(-1) = -1 +3 = 2
To determine the extreme value for a function, minimum or maximum, we have to differentiate the function,which, in this case, is:
f'(x)= 3x^2 - 3
Now, we'll calculate the root of the first derivative, this value being the value for the function f has an extreme value.
3x^2 - 3 = 0
We'll solve the difference of squares:
We'll put each factor of the product as being zero.
So, the extreme values of the function are:
f(1) = 1^3-3
f(1) = -2
f(-1) = -1+3
f(-1) = 2
To find the extreme values of f(x) =x^3-3x.
f(x) = x^3-3x = x(x^2-1) = x(x^2-1) = (x+1)(x)(x-1).
For x>1, f(x) positive is increasing and goes unbounded as x-->infinity.
For x=1, f(x) = 0.
For 0<x<1, f(x) is <0 or negative,
For x= 0, f(x) = 0.
For -1< x < 0, f(x) > 0 or positive.
For x =-1 , f(x) = 0 and
For X < -1, f(x) is negative and goes to minus infinity.
The local extreme values of the function is got by differentiating f(x) with respect to x setting the f'(x) = 0 and solving for x=c. Again if f"(c) < 0, then f(c) is mximum for x= c. If f"(c) > 0, then f'(c) is the minimum for x = c.
f(x) = x^3-3x. Differentiating both sides, setting f'(x) = 0,
3x^2-3 = 0. Or
x^2-1 = 0. Or
x^2 = 1. Or x = 1 Or x = -1.
f''(x) = (3x^2-3)' = 6x.
f''(1) =6*1 is positive. So for x =1. f(1) = 1^3-3*1 = -3 is a local minimum.
f"(-1) = 6(-1) = -1 . So f(-1) = 3(-1)^2-3 = 0 is the local maximum.