Find tangent that is perpendicular using differentation formula?

Question

Here is problem in original form.

Find the point(s) where the tangent line to f(x)=((x^2)-1)(x+7) is perpendicular to the line y=(1/2)x+7

Here is what I have done:

y=(1/2)x+7 so m=1/2 and perpendicular slope is -2

I do not know where to go on because I am very, very tired...I have slept only 7 hours in the past 48 hours, so please help.

Answer 1

Find the derivative and put it =-2
f´(x) = 2x(x+7) +x^2-1 = 3x^2 +14x -1=-2
3x^2+14x+1=0 No real solutions

Answer 2

f '(x) = 2x - 1
slope of line = 1/2
perpendicular to line: slope =2 . . . . equating
2x -1 = 2
x = 3/2
y = 1/2 (3/2) + 7 = 3/4 + 7 = 31/3

equation of line
1/2 = (y-31/3)/(x-3/2)
x - 32/ 2 = 2 (y - 31/3)
(2x -32)/2 = 2 (3y - 31)/3
3 (2x -32) = 4 (3y - 31)
6x - 96 = 12 y - 124
6x - 12y + 28 = 0
3x - 6 y + 14 = 0

Answer 3

f(x)=(x²-1)(x+7) = x³ + 7x² - x - 7

f'(x) = 3x² + 14x - 1

3x² + 14x - 1 = -2

3x² + 14x + 1 = 0

x = (-14 ± √(196 - 12))/6

= (-7 ± √46)/3

f(x) = (x²-1)(x+7)

= ((-7 + √46)/3)² - 1)(((-7 + √46)/3) + 7)

or ((-7 - √46)/3)² - 1)(((-7 - √46)/3) + 7)

= (560 + 72√46)/27 or (560 - 72√46)/27

Solutions are
((-7 + √46)/3),(560 + 72√46)/27))
or
((-7 - √46)/3),(560 - 72√46)/27))