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Albert1
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$f(x)=\sqrt {8x-x^2}-\sqrt{14x-x^2-48}$
find :$min(f(x))$ and $max(f(x))$
find :$min(f(x))$ and $max(f(x))$
max=$2\sqrt 3$ correctjacks 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 $x=7$
The purpose of finding the min and max of a function is to determine the lowest and highest values that the function can take on within a given interval. This information can be useful in understanding the behavior and characteristics of the function.
To find the min and max of a function, you can use the first derivative test or the second derivative test. These methods involve taking the derivative of the function, setting it equal to zero, and solving for the critical points. Then, you can plug these critical points into the original function to determine which one is the min and which one is the max.
Yes, a function can have more than one min or max. This can happen when the function is not continuous or when there are multiple critical points that have the same value.
A local min/max is the lowest/highest value of a function within a specific interval, while a global min/max is the lowest/highest value of a function within the entire domain. A global min/max is also referred to as the absolute min/max.
Finding the min and max of a function can be applied in various fields such as economics, engineering, and physics. For example, in economics, finding the minimum cost or maximum profit function can help businesses make strategic decisions. In engineering, finding the maximum load a structure can withstand can ensure the safety and stability of a building. In physics, finding the minimum/maximum values of a function can help determine the maximum/minimum velocity or acceleration of an object.