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

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SUMMARY

The discussion focuses on the function $f(x)=\sqrt{8x-x^2}-\sqrt{14x-x^2-48}$. The maximum value of this function is definitively established as $2\sqrt{3}$, while the minimum value is confirmed to be $0$. Participants utilized algebraic manipulation and properties of square roots to derive these values, ensuring clarity in the solution process.

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

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