Given f(x)=1/3(4-x)^2 , what is the value of f(16)?
I think this is (1/3) (4 - x)^2
So
f(16) = (1/3)(4 - 16)^2 = (1/3)(-12)^2 = (1/3)(144) = 48
\(f(x)=\frac{1}{3(4-x)^2}\)
\(\frac{1}{3(4-16)^2}=\frac{1}{3\times 144}=\frac{1}{432}\)
You are very welcome!
:P
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