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I am told to find the absolute extrema of

$$h(x) = \frac{8+x}{8-x},[4,6]$$

So I obtain the derivative of

$$\frac{16}{(8-x)^2}$$

The trouble I am having is trying to determine the critical points. I know that a critical point is found where the derivative is equal to zero or does not exist.

So my assumption would be that the critical point is

  1. $\frac{16}{(8-8)^2}$, i.e. x = 8

Looking for someone to shed some light on this. Would there be no critical points because the function is discontinuous (I think) and therefore I would just test the endpoints?

Thank you for any help!

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Hints:

$$\frac{8+x}{8-x}=1+\frac{2x}{8-x}=1+\frac{2}{\frac8x-1},$$

clearly, it is an incresing function at $[4,6]$.

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  • $\begingroup$ Thank you! Since there is no sign change there is no critical point therefore only the endpoints should be checked. $\endgroup$ – mrybak834 Oct 21 '14 at 8:49
  • $\begingroup$ You are welcome! $\endgroup$ – Paul Oct 21 '14 at 8:53
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Since you obtained that the derivative is $$\dfrac{16}{(8-x)^2}>0$$ it is straightforward that your function is monotone increasing in the given interval (positive derivative for all $x \in [4,6]$). Therefore in order to find it's extrema you should look at the boundaries of the interval.

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  • $\begingroup$ Thank you very much Stefanos! Understood! $\endgroup$ – mrybak834 Oct 21 '14 at 8:49
  • $\begingroup$ I corrected it, but the point is the same. You are welcome. $\endgroup$ – Jimmy R. Oct 21 '14 at 8:50

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