# Finding the points of the curve where the tangent line is horizontal

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?

• You don't have to, since $\left (\ln |x| \right )' = \frac 1x$ – Kaster Mar 1 '13 at 23:45
• @Kev: The $=$ should be a $-$. Now, you just need to find the zeros of the derivative. In other words, how will that fractional derivative be equal to zero? – Amzoti Mar 1 '13 at 23:55
• – Daryl Mar 1 '13 at 23:57
• @Daryl not too helpful though, you missed part left from the vertical asymptote. This should do it :) – Kaster Mar 2 '13 at 0:36

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):

• After taking the derivative and bringing everything under the same denominator I get (x-2) +(x-2)^2 +12 in the numerator. I expand this and arrive at x^2 - 3x + 14. Yours definitely looks more plausible but I just can't see how you arrive at it. – Kev Mar 2 '13 at 0:09
• After derivating your function, the last term before finding a common denominator, is $\dfrac {-12}{(x-2)^2}$, so the 12 in the numerator should be -12 (recall, the exponent on the $(x - 1)$ in the last term is $-1$, so deriving $\dfrac{12}{(x-2)} = \dfrac{-12}{(x - 2)^2}$ – Namaste Mar 2 '13 at 0:17
• Ah ok, I see that I mixed up the sign on the 12. So I would get two points, where x is 5 and -2. – Kev Mar 2 '13 at 0:26
• Exactly. Just be sure to find the corresponding y-values for those points, using the original equation. – Namaste Mar 2 '13 at 0:37
• Yeap, got it. Thanks a lot. – Kev Mar 2 '13 at 0:43