# For f(x) = -x^3 - 6x^2 - 9x - 2 determine all the critical pointstest each interval and use the first and second derivative test to determine where the graph is increasing and which critical...

For f(x) = -x^3 - 6x^2 - 9x - 2
determine all the critical points

test each interval and use the first and second derivative test to determine where the graph is increasing and which critical points are max and min values.

sciencesolve | Certified Educator

Determine the first derivative of function to check if there are extreme points.

`f'(x) = -3x^2 - 12x - 9`

Solve the equation `-3x^2 - 12x - 9 = 0` .

Dividing the equation by -3 yields: `x^2 + 4x + 3 = 0`

Use quadratic formula to determine the roots of the equation  `x^2 + 4x + 3 = 0` .

`x_(1,2) = (-4+-sqrt(16-12))/2`

`` `x_(1,2) = (-4+-sqrt(4))/2 =gt x_(1,2) = (-4+-2)/2`

`` `x_1`  = -1 and `x_2`  = -3

Let's check how the sign of the function f'(x) changes over the intervals`(-oo ; -3)`  ; (-3 ; -1) and `(-1 ; oo).`

Consider a value of x included in `(-oo ; -3) ` and calculate f'(x) at this value.

`x = -4 =gt f'(-4) = (-4)^2 + 4*(-4)+ 3 = 16- 16 + 3 = 3 gt 0`

The function f'(x) has positive values if x  `in` `(-oo ; -3), ` therefore the function f(x) increases in`(-oo ; -3).`

Consider a value of x included in ( -3;-1) and calculate f'(x) at this value.

`x = -2 =gt f'(-2) = (-2)^2 + 4*(-2)+ 3 =4-8 + 3 =-1lt 0`

The function f'(x) has negative values if x is in ( -3;-1), therefore the function decreases in ( -3;-1).

If the function increases in `(-oo ; -3)`  and it decreases in ( -3;-1) => f(-3) is an extreme points of local maximum.

f(-3) =27 - 54 + 27 - 2 = -2

Consider a value of x included in`(-1 ; oo)`  and calculate f'(x) at this value.

`x = 0=gt f'(0) = (0)^2 + 4*(0)+ 3 = 3gt 0`

The function f'(x) has positive values if x is in `(-1 ; oo), ` therefore the function f(x) increases in `(-1 ; oo).`

If the function decreases in ( -3;-1) and it increases in  `(-1 ; oo)`  => f(-1) is an extreme points of local minimum.

f(-1) = 1-6+9-2 = 2

Calculate the second derivative of function to check if there are inflection points.

f"(x) = 2x+4

Calculate the roots of f"(x) = 0 => 2x+4=0 => x+2=0 => x = -2

The function f(x) has an inflection point at x=-2.

f(-2) = 8-24+18-2=0

The function has a point of minimum at (-1;2), a point of maximum at (-3,-2) and a point of inflection at (-2;0).