f(x)= x^3 + ax^2 + bx , has critical #'s at x= -1 and x= 2... find a and b exact values?

Question

best answer to first correct response

Answer 1

f '(x) = 3x^2+2ax+b
f '(-1) = 3 -2a+b = 0
f '(2) = 12 +4a +b = 0
6-4a +2b = 0
18 +3b = 0
b = -6
3-2a -6=0
-2a =3
a= -1.5
So equation is x^3 -1.5x^2 -6x

Answer 2

I am assuming that critical numbers occur when y = 0, therefore you just have to plug in your 2 x's and set the equations equal to 0.

0 = (-1)^3 + a(-1)^2 +b(-1) = -1 + a - b
and
0 = 2^3 + a*2^2 + b*2 = 8 + 4a + 2b

Now you have 2 simultaneous equations to solve for:

0 = -1 + a - b
0 = 8 + 4a + 2b

rearranging

1 = a - b
-8 = 4a + 2b

Multiply the first equation by 2 and add the equations together to isolate a:

2 = 2a - 2b
-8 = 4a + 2b
------------------
-6 = 6a
a = -1

Use the first equation to solve for b

1 = -1 - b
b = -2

Answer 3

f'(x) = 3x^2 +2ax +b =0

3(x+1)(x-2) = 3( x^2 -x -2) = 3x^2 -3x -6 =3x^2 +2ax +b;
2a =-3 and b =-6
a =-1.5

f(x) = x^3 - 1.5x^2 -6x

f(-1) = 3.5
f(2) = -10

Answer 4

a=3
b=2

you just factor the polynomial into parts and reverse engineer the quadrant possiblilites.

Answer 5

huuuuuuuu what the hell. thats easy it = 74652.764 thats simple