If $F (x) = \sqrt{64} - x^{2}$ $F(x) = sqrt 64 - x^2$ , what is f(x)?

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Answer 1 · 2015-01-27T00:16:14

It your function is as shown, it would be $f (x) = - 2 x$ $f(x)=-2x$

Because $\sqrt[2]{64} = 8 \to F (x) = 8 - x^{2}$ $root 2 64 = 8->F(x)=8-x^2$

But I think you meant $\sqrt[2]{64 - x^{2}} = \sqrt[2]{(8 + x) (8 - x)}$ $root 2 (64-x^2)= root 2 ((8+x)(8-x)$

Please confirm in a comment, before I go any further.

BTW :
We use the $f(x)->f'(x)$ notation.
I hardly ever see the $F(x)->f(x)$

If F(x) = sqrt 64 - x^2F(x)=√64−x2F(x) = sqrt 64 - x^2, what is f(x)?