Reversal Point and Tangent line

Question

Given is the function $\frac{1}{4}x^4-x^3+2x$, the task is to find the the turning points and then the tangent line to the left-most turning point.

Deriving the functin we get $x^3 -3x^2 + 2x$ and with $x^3 -3x^2 + 2x = (x-1)(x^2-2x-2)$ the solutions to $(x-1)(x^2-2x-2) = 0$ are $x_1 =1$, $x_2 = 1 + \sqrt{3}$ and $x_3 = 1 - \sqrt{3}$. The leftmost turning point is $x_3$. Regarding the tangentline at this point: As far as I understand, the slope of the tangent corresponds to the slope of the original function at this point, which would be zero here since at a minima the function has slope zero. Is hence the equation for tangent simply $-1 = 0*(1-\sqrt{3})-1$?

André Nicolas · Accepted Answer · 2013-05-08T14:53:36.960

0

The tangent line has slope $0$, so has equation of shape $y=c$. To find the value of $c$, substitute $1-\sqrt{3}$ in the expression for our function. If we want to simplify, it is a relatively messy process, since we are substituting into a degree $4$ polynomial.

Remark: Ther is a typo in the OP, $x^3-3x^2+2$ was intended. But that is what you actually worked with, so the calculations are correct. Things only break down at the end, at the equation of tangent line phase.

edited May 08 '13 at 14:53

answered May 08 '13 at 14:46

André Nicolas

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Hi! Not sure..so I did in fact substituted $1 - \sqrt{3}$, and hence I just got c = -1 since as you also mention the slope is 0. So how is then my equation wrong? :-) thank you – May 08 '13 at 15:17
Your equation has no $y$ in it. – André Nicolas May 08 '13 at 15:19
I see, sorry. So then y = -1 would be the tangent. – May 08 '13 at 15:20
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Yes, that's the tangent line. (I checked the correctness of the substitution.) – André Nicolas May 08 '13 at 15:23

Reversal Point and Tangent line

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