Find x-coordinate, given the derivative and horizontal tangent.

Question

dy/dx = (2x-y)/(x+y) Horizontal tangent at y=8.

How does one determine the x-coordinate of the point? I'm not sure.

What's the value of a derivative in case of a horizontal tangent? — imranfat, Aug 29 '14 at 15:22

score 1 · Answer 1 · answered Aug 29 '14 at 15:20

1

So at $y=8,$ the gradient is same as that of the x axis which is $0$

$$\frac{dy}{dx}_{(\text{ at }y=8)}=\frac{2x-8}{x+8}$$

answered Aug 29 '14 at 15:20

lab bhattacharjee

274,582

score 0 · Answer 2 · answered Aug 29 '14 at 15:40

The tangent is horizontal at $y=8$, so the gradient must be nil, so $\frac{2x-8}{x+8}=0$ and hence $x=4$. So the curve passes through the point (4,8).

Integrating the equation given we get a quadratic in $y$. Solving we get $y=-x\pm\sqrt{3x^2+k}$. Since it passes through (4,8) we get $k=96$ and $y=-x+\sqrt{3x^2+96}$.

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Find x-coordinate, given the derivative and horizontal tangent.

2 Answers2