To find the extreme values of x^2+6x-6.
We write x^2+6x-6 as (x^2+6x+3^2) -3^2 -6 = 0. We added here 3^2 and subtracted.
Therefore x^2+6x-6 = (x+3)^2 -9-5
x^2+6x-6 = (x+3)^2 -15.
x^2+6x-6 > A perfect square (x+3)^2 -15.
Therefore x^2+6x -6 is always greater than -15, as (x+3)^2 > 0 for all x and is minimum when x+3 = 0 or x = -3.
Therefore x^2+6x-6 has the minimum (or extreme value) -15 when x = -3.
The extreme values of a function are the points where the value of the differential of the function is equal to zero. Here we have the function f(x) = x^2 + 6x - 6 .
Differentiating f(x) = x^2 + 6x - 6, we use the rule that the derivative of x^n is n*x^(n-1)
Therefore f'(x) = 2x + 6
Equating f'(x) to 0
=> 2x +6 = 0
=> x = -3
Now substituting x= -3 in f(x) = x^2 + 6x -6 = 9 - 18 -6 = -15
Therefore the extreme value is -15.
Since the expression of the function is a quadratic, we'll consider the vertex of the parable as the extreme point.
Since the coefficient of x^2 is positive, the vertex is a minimum point.
f(x) = x^2 + 6x - 6
The minimum value is V(-b/2a , -delta/4a)
We'll identify the coefficients a,b,c:
a = 1
b = 6
c = -6
delta = b^2 - 4ac
delta = 36 + 24
delta = 60
The vertex V has the following coordinates:
xV = -b/2a
xV = -6 / 2*1
xV = -3
yV = -60/4*1
yV = -15
The minimum point of the function is V(-3 , -15).
Another method to determine the extreme point of a function, is the first derivative test.
We'll calculate the first derivative of the function:
f'(x) = (x^2 + 6x - 6)'
f'(x) = 2x + 6
Now, we'll compute the roots of f'(x):
2x + 6 = 0
We'll subtract 6 both sides:
2x = -6
x = -3
The roots of the first derivative represent the extreme points of the function.
We'll calculate the extreme point by substituting x by -3 in the original function:
f(-3) = (-3)^2 + 6(-3) - 6
f(-3) = 9 - 18 - 6
f(-3) = -15
The extreme point has the coordinates: (-3 , -15).