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Determine the input that yields the minimum value for the function f(x)=square root of (x^2-8x+41) I just don't know how to find the input when it is the square root of
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y = sqrt(x^2-8x+41) = sqrt((x-4)^2 + 25) to get the minimum value, it should when x=4
so, just put x=4 to that function what is the value of y will u get ?
y=f(x)
any way u can use derivative to get it, but i think thought one like on
Do you know calculus?
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