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Find minimum of function $f(x) = \frac{\left| x-12\right| }{5}+\frac{\sqrt{x^2+25}}{3}$

I tried compute min using by definition of abs function. I consider two cases:

  1. when $x > 12$ we have: $f_{1}(x) = \frac{x-12}{5}+\frac{\sqrt{x^2+25}}{3}$ using standard method $f_{1}'(x)=0$ for $x_{0}=-15/4$ but $x_{0}$ is not in $ [12, \infty]$
  2. when $x \leq 12$ we have $f_{2}(x) = \frac{-x+12}{5}+\frac{\sqrt{x^2+25}}{3}$ similarly $f_{2}'(x) = 0$, and $x = 15/4$

So using observation from this method I computed minimum of $f(15/4) = 56/15$.

Does it be a correct way?

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  • $\begingroup$ Your method appears to be correct. $\endgroup$
    – Allawonder
    Aug 27, 2019 at 22:05
  • $\begingroup$ 1) can be reasoned more simply that $f_1(x)$ is strictly increasing on $x > 12$ so there can't be any minimum there. $\endgroup$
    – fleablood
    Aug 27, 2019 at 23:23
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    $\begingroup$ For good measure, I would also compute the value at the boundary $x=12$ to see how it compares. To see why this is necessary, try using your method to find the minimum of the function $f(x)=|x|$. $\endgroup$
    – Théophile
    Aug 28, 2019 at 0:55

2 Answers 2

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We can use C-S and the triangle inequality:

$$\frac{|x-12|}{5}+\frac{\sqrt{x^2+25}}{3}=\frac{|x-12|}{5}+\frac{\sqrt{(3^2+4^2)(x^2+5^2)}}{15}\geq$$ $$\geq \frac{|x-12|}{5}+\frac{|3x+20|}{15}=\left|\frac{12}{5}-\frac{x}{5}\right|+\left|\frac{4}{3}+\frac{x}{5}\right|\geq $$ $$\geq \left|\frac{12}{5}-\frac{x}{5}+\frac{4}{3}+\frac{x}{5}\right|=\frac{56}{15}.$$ The equality occurs for $(3,4)||(x,5)$ and $\left(\frac{12}{5}-\frac{x}{5}\right)\left(\frac{4}{3}+\frac{x}{5}\right)\geq0,$ which gives $x=\frac{15}{4},$

which says that we got a minimal value.

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Your reasoning is spot on!

Indeed, since both terms are never negative, it is certain that there is a minimum. But you have exhibited the only possibility. It follows that that is indeed the minimum value of the function.

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    $\begingroup$ $\dfrac{|x-12|}5\ge0$ and $\dfrac{\sqrt{x^2+25}}3\ge5,$ but that does not mean the minimum of their sum is ever as low as $0+5$ $\endgroup$ Aug 27, 2019 at 22:10
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    $\begingroup$ @J.W.Tanner I realised this, too. Thanks, though. Accordingly edited. $\endgroup$
    – Allawonder
    Aug 27, 2019 at 22:16

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