MHB What are the minimum and maximum values of the given function?

[Albert1]

$f(x)=\sqrt {8x-x^2}-\sqrt{14x-x^2-48}$
find :$min(f(x))$ and $max(f(x))$

 

[juantheron]

Spoiler

Given $f(x) = \sqrt{8x-x^2} - \sqrt{14x-x^2-48} = \sqrt{16-(x-4)^2} - \sqrt{1-(x-7)^2}$

Now Drawing Two half Circle,

$y_{1} = \sqrt{16-(x-4)^2}$ and $y_{2} = \sqrt{1-(x-7)^2}$

I am getting ..

Max. of $(y_{1}-y_{2}) = 2\sqrt{3}$ at $x=6$ and Min. of $(y_{1}-y_{2}) = 0$ at $x=8$

Edited it.

 

[Albert1]

Spoiler

Given $f(x) = \sqrt{8x-x^2} - \sqrt{14x-x^2-48} = \sqrt{16-(x-4)^2} - \sqrt{1-(x-7)^2}$

Now Drawing Two half Circle,

$y_{1} = \sqrt{16-(x-4)^2}$ and $y_{2} = \sqrt{1-(x-7)^2}$

I am getting ..

Max. of $(y_{1}-y_{2}) = 2\sqrt{3}$ at $x=6$ and Min. of $(y_{1}-y_{2}) = \sqrt{7}-1$ at $x=7$

max=$2\sqrt 3$ correct
min :$0$ correct