Asked by James
Find the equation of each normal of the function f(x) = 1/3 * x ^ 3 + x ^ 2 + x - 1/8 which is parallel to the line
y=- 1 4 x+ 1 3 .
y=- 1 4 x+ 1 3 .
Answers
Answered by
oobleck
AAAaannndd the bot gets it wrong yet again!
x^2 + 2x + 1 = -1/14
(x+1)^2 = -1/14
There are no real solutions to this, since a square cannot be negative.
Of course, since the tangent line must be parallel to y = -14x+13, the slope of the normal is 1/14.
x^2 + x + 1 = 1/14
x = -1 ± 1/√14
x^2 + 2x + 1 = -1/14
(x+1)^2 = -1/14
There are no real solutions to this, since a square cannot be negative.
Of course, since the tangent line must be parallel to y = -14x+13, the slope of the normal is 1/14.
x^2 + x + 1 = 1/14
x = -1 ± 1/√14
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