Given $y = x^3 − 2x$ for $x \geq 0$, find the equation of the tangent line to $y$ where the absolute value of the slope is minimized.

Question

I tried finding the derivative of this, and promptly got $y=$ about $0.816$, but I have no idea how to put that into equation form or if I'm even correct.

Welcome to Mathematics Stack Exchange. What do you get for the slope of the tangent line at $x$? — J. W. Tanner, Jul 19 '20 at 03:43

Math Lover · Accepted Answer · 2020-07-19T04:44:44.633

1

I assume you already got $y' = 3x^2 - 2$. This is the slope of the curve $y$.

As per the question, you need to find absolute minimum of the slope which in
this case is $0$ for $x = \sqrt{\frac{2}{3}}$, (where $x \ge 0)$.

Substituting x in your curve, you get the equation of the tangent line as the slope is zero.

$y + \frac{4\sqrt2}{3\sqrt3} = 0$

edited Jul 19 '20 at 04:44

answered Jul 19 '20 at 04:07

Math Lover

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Given $y = x^3 − 2x$ for $x \geq 0$, find the equation of the tangent line to $y$ where the absolute value of the slope is minimized.

1 Answers1