Finding maximum value for a function I was working on this question to find the following function's maximum value.Let $$y=f(x)={{(\sqrt{-3+4x-x^2}+4)}}^2 + (x-5)^2$$ where $$1 \le x \le 3$$.I have to find it's maximum value.
I tried by simple method of taking $$\frac{dy}{dx}=0$$ and got the answer.But this path was pretty lenghty and tiresome.Is there any short method possible for this?

You have simply$$y=(\sqrt{-3+4x-x^2}+4)^2 + (x-5)^2 \\
y=(\sqrt{1-(x-2)^2}+4)^2 + (x-5)^2 \\
y'=2(\sqrt{1-(x-2)^2}+4).\frac{-2(x-2)}{2\sqrt{1-(x-2)^2}} + 2(x-5)\\
y'=(\sqrt{1-(x-2)^2}+4).\frac{-2(x-2)}{\sqrt{1-(x-2)^2}} + 2(x-5)\\
y'=\frac{-2(x-2)\sqrt{1-(x-2)^2}-8x+16+2(x-5)\sqrt{1-(x-2)^2}}{\sqrt{1-(x-2)^2}}\\
y'=\frac{\sqrt{1-(x-2)^2}-8(x-2)}{\sqrt{1-(x-2)^2}}\\$$
$y'=0 \to \sqrt{1-(x-2)^2}-8(x-2)=0$ then
$$1-(x-2)^2=64(x-2)^2\\
(x-2)^2=\frac{1}{65}\\
x=2+\frac{1}{\sqrt{65}}\\
x=2-\frac{1}{\sqrt{65}}$$
You have to compare the values of $f$ at these two values and at $x=1$ and at $x=3$. I hope it helps you.
