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The curve $xy =1$ has a slope that is negative everywhere in the first quadrant. This is apparent both visually and algebraically ($dy/dx = -y/x$).

Its reflection in the line $x=4$ is the curve $y(8-x) = 1$ (Reflection of rectangular hyperbola in vertical line)

When I plot the curve using R it seems as if the slope of the reflected curve is positive everywhere in the first quadrant. But algebraically the slope $dy/dx$ is given by $y/(8-x)$ which indicates that when x is greater than 8 the slope is negative.

I can't believe that my eyes are deceiving me. Is the equation for the slope correct?

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  • $\begingroup$ When $x$ is greater than $8$, the slope is indeed negative. What is the problem? $\endgroup$
    – DHMO
    Apr 18, 2017 at 7:56
  • $\begingroup$ I think $y=\frac{1}{8-x}\implies \frac{dy}{dx}=\frac{1}{(8-x)^2}>0$ $\endgroup$
    – Lazy Lee
    Apr 18, 2017 at 8:01
  • $\begingroup$ @LazyLee chain rule, 1/u and 8-x =u $\endgroup$
    – Xetrov
    Apr 18, 2017 at 8:12

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When $x>8$, I agree that $8-x<0$ but also notice that since $y=\dfrac{1}{8-x}$ , $y<0 ; ~\text{when}~ x>8$,

Hence the slope = $\dfrac{y}{8-x}= \dfrac{\text{negative}}{\text{negative}}=\boxed{\text{positive}}$ as expected by you.

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    $\begingroup$ Thanks a lot. But after your answer I realise there's more to it. $x =8$ is the asymptote of $y(8-x) = 1$. When y is negative we are dealing with the reflection of the part of the rectangular hyperbola that lies in the third quadrant. $\endgroup$
    – phil342
    Apr 18, 2017 at 11:02

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