Matematické Fórum

stuart clark · 10. 12. 2013 05:05

Minimum value of $f(x) = \left|\sqrt{x^2-8x+52} - \sqrt{x^2-4x+8}\right|.$ where $x\in \mathbb{R}$

Brano · 10. 12. 2013 14:36

↑ stuart clark:
$|\text{anything}|\ge 0$
so it is sufficient to find $x$ where the equality holds, i.e. solve
$\sqrt{x^2-8x+52} = \sqrt{x^2-4x+8}$

stuart clark · 16. 12. 2013 04:19

Thanks Brono actually here we have to calculate Range of $f(x)$ .

Thanks

Brano · 17. 12. 2013 12:36 — Editoval Brano (17. 12. 2013 12:48)

↑ stuart clark:
candidates for points of global "maximum" are stationaty points of $\sqrt{x^2-8x+52} - \sqrt{x^2-4x+8}$ and $\pm\infty$ - to be more precise in the case of $\pm\infty$ you have to compute limits of $f$ and it would be supremum and the range would be half open interval.
But in this case the value at the stationaty point is greater, see
http://www.wolframalpha.com/input/?i=|s … %3D-15..40

Matematické Fórum

#1 10. 12. 2013 05:05

Minimum value

#2 10. 12. 2013 14:36

Re: Minimum value

#3 16. 12. 2013 04:19

Re: Minimum value

#4 17. 12. 2013 12:36 — Editoval Brano (17. 12. 2013 12:48)

Re: Minimum value

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