[Math] Finding the points of the curve where the tangent line is horizontal

absolute valuederivatives

The curve given is $\displaystyle y = \ln|x-2| + x + \frac{12}{x-2}$.

Find the points of the curve where the tangent line is horizontal.

My first stumbling block is the absolute value function. I know the value within it would be $\ge 0$ when $x \ge 2$ and it would be $\lt 0$ when $x \lt 2$.

How do I remove the absolute value brackets for differentiation?

EDIT: After being told that I don't need to remove the brackets, I eventually obtained $(x^2-3x+14)/(x-2)^2 = 0.$ At this point do I simply factorize the numerator to obtain the points?

Best Answer

Set the numerator of the derivative equal to zero (we want the slope of the curve = slope of the horizontal tangent line = $0$) and solve for $x$; those solutions will give you the $x$-coordinates of the points on the curve where the tangent line is horizontal.

  • Your derivation is slightly off: The numerator should be $x^2 - 3x - 10$, so put $$x^2 - 3x - 10 = 0$$ factor, and solve.

    (See comment below: you forgot to multiply $12$ in the last term by $-1$ when differentiating that third term.)

I take it you know how to proceed from here: Go back to the original equation to find the points of interest: $(5, y_1),\; (-2, y_2)$.

Here's a helpful plot of your function to "confirm" visually where you'd find the points where the tangent lines are horizontal (compliments of Worlfram Alpha):

enter image description here

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