Finding points on a curve where the tangent slope = 6

[dysania]

Homework Statement

Find the x values of the points on the curve f(x) = (x^3) + (3x^2) + 3x + 6 where the tangent has a slope equal to m = 6

Homework Equations

The Attempt at a Solution

This question was on an online quiz for my into calculus course... can't seem to wrap my head around it.

I differentiated:
f'(x) = 3(x^2) + 6x + 3

Set it equal to 6:

3(x^2) + 6x + 3 = 6

Then attempted to factor and solve for x, in different ways:

3(x^2) + 3x + 3x + 3 = 6
3x(x+1) + 3(x+1) = 6
(3x+3)(x+1) = 6
3x+3=6
x=(6-3)/3
x=1
x+1=6
x=5

I also tried:
3(x^2) + 6x + 3 = 6
3((x^2) + 2x + 1) = 6
3(x+1)(x+1) = 6
(x+1)(x+1) = 6/3
x+1 = 2
x = 1

Needless to say I got the problem wrong... the correct answers were given: -1 + sqrt(2), and -1 - sqrt(2).

I would appreciate any kind of help or pointers... I just can't seem to get this straight. What am I missing??

 

[futurebird]

I have no idea what you are doing with that factoring! You can only factor when one side is set equal to zero. (Remember we can solve by factoring because ab=0 implies that a=0 or b=0 ... but ab=6 will NOT imply a=6 or b=6!)

You have:

6 = 3x^2 +6x +3

Set it equal to zero, like this:

0 = 3x^2 +6x - 3

But this will not factor... so, use the quadratic formula.

 

[dysania]

I have no idea what you are doing with that factoring! You can only factor when one side is set equal to zero. (Remember we can solve by factoring because ab=0 implies that a=0 or b=0 ... but ab=6 will NOT imply a=6 or b=6!)

You have:

6 = 3x^2 +6x +3

Set it equal to zero, like this:

0 = 3x^2 +6x - 3

But this will not factor... so, use the quadratic formula.

Thank you, I honestly had no idea what I was doing either. Appreciate the help.