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
 A: Let $x = 2+\cos \theta$ where $0\leq \theta \leq \pi$. Then the equation is equivalent to 
$$
(\sin\theta + 4)^{2} + (\cos\theta-3)^{2} = 26 + 8\sin\theta - 6\cos\theta = 26 + 10\sin(\theta+\alpha)
$$
where $-\pi/2 < \alpha <0$ satisfies $\tan \alpha = -3/4$. This attains minimum at $\theta = \pi/2 - \alpha$, which corresponds to $x=7/5$.  
A: With a bit of arrangements I did on paper, you could show that 
\begin{equation}
 f'(x) 
 =
 -\dfrac{6\sqrt{-x^2+4x-3}+8x-16}{\sqrt{-x^2+4x-3}}
\end{equation}
which is zero when
\begin{equation}
 6\sqrt{-x^2+4x-3}= -8x+16
\end{equation}
square both sides
\begin{equation}
 -36x^2+ 144x - 108 = 64x^2 -256x + 256
\end{equation}
which is 
\begin{equation}
 100x^2 -400x +364 = 0
\end{equation}
Roots are
\begin{equation}
 x_{1,2} 
 =
 \frac{400 \pm 120}{200}
\end{equation}
i.e.
\begin{align}
 x_1 &= 2.6 \\
 x_2 &= 1.4
\end{align}
Either do a double derivative (which requires much time) to show that $f(x_2)$ is a maximum, or wiggle around $x_2$ if you're in a hurry.
A: 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 $x=t+2=-\frac35+2=\frac75$.
