<s>[INST] maximum value of expression $(\sqrt{-3+4x-x^2}+4)^2+(x-5)^2$ maximum value of $\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
what i try
$\displaystyle -3+4x-x^2+16+8\sqrt{-3+4x-x^2}+x^2+25-10x$
$\displaystyle -6x+38+8\sqrt{-3+4x-x^2}$
using derivative it is very lengthy
help me how to solve, thanks in advance
 [/INST] Given: 

$\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$

Denote $x=t+2$, then:
$$f(t)=\bigg(\sqrt{1-t^2}+4\bigg)^2+\bigg(t-3\bigg)^2\;\forall\;t\in[-1\;,1]\\
f'(t)=2\bigg(\sqrt{1-t^2}+4\bigg)\cdot \frac{-t}{\sqrt{1-t^2}}+2(t-3)=0 \Rightarrow\\
-2t-\frac{8t}{\sqrt{1-t^2}}+2t-6=0 \Rightarrow \\
4t=-3\sqrt{1-t^2} \stackrel{t<0}{\Rightarrow} \\
16t^2=9-9t^2 \Rightarrow \\
t=-\frac35.$$
Hence, for $f(t)$:
$$f(-1)=32\\
f(1)=20\\
f(-\frac35)=46.08 \ \text{(max)}.$$
It means $x=t+2=-\frac35+2=\frac75$.
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