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

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The function f(x) = √(8x - x²) - √(14x - x² - 48) is analyzed for its minimum and maximum values. The maximum value of the function is confirmed to be 2√3. The minimum value is established as 0. These results indicate the function's behavior within its defined domain. The findings provide a clear understanding of the function's extremities.
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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:

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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