X's when derivative = 2?

Question

I'm finding the x-coordinates of x^4-3x^2+2 when its derivative is 2. I put the derivative = to 2, factored, and got (x+1)(4x^2-4x-2)=0. Now what? I CANNOT USE A CALCULATOR (take home quiz)

Answer 1

Hi,
Good work, you're almost there. The derivative is 2 at three different values of x. So, you have one:
x+1=0
x=-1
Now, for the other two.
4x²-4x-2 =0
2(2x²-2x-1) = 0
This is not easily factorized, so we use the quadratic formula:
x=(-B +- sqrt(B²-4AC))/(2A)
x=(-(-2)+-sqrt ((-2)²-4(2)(-1))/4
x= (2+-sqrt(12))/4
x=(2+-sqrt(4*3))/4
x = (1+sqrt(3))/2
So, those plus -1 are the x-values where the derivative is two.
If you should need the corresponding y-values, just substitute those x-values in the original function and solve for y. I know you wouldn't want me to deprive you of doing the arithmetic for that. :-)l

Hope this helps.
FE

Answer 2

x^4-3x^2+2

derivative = 4x^3 -6x =2 or 4x^3-6x -2 =0
you already found that x =-1 is a root

we divide 4x^3-6x -2 by (x+1) to get
4x^2-4x-2=0
Discriminant is 16 +4(2)(4) = 16+32 =48
x1 = {4 -sqrt(48)}/8 = (1-sqrt(3))/2
x2 = {4 +sqrt(48)}/8 = (1+sqrt(3))/2

Answer 3

dy/dx=4x^3-6x=2
4x^3-6x-2=0
(x+1)(4x^2-4x-2)=0
if x+1=0, x=-1
if 4x^2-4x-2=0
ax^2+bx+c=0
x=[-b+sqrt(b^2-4ac)]/2a
x=[-b-sqrt(b^2-4ac)]/2a
a=4 b=-4 c=-2
solving x=1.366 , -0.366 (calculator)