F(x)=root of (x^2+16) and (f(k))^2=20,k>0.what is the value of f(k+1)?

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Answer 1 · 2018-05-27T03:52:14

$f (k + 1) = 5$ $f(k + 1) = 5$

We have:

${(f (x))}^{2} = {(\sqrt{x^{2} + 16})}^{2} = x^{2} + 16$ $(f(x))^2 = (sqrt(x^2 + 16))^2 = x^2 + 16$

We now can solve for $k$ $k$ .

${(f (k))}^{2} = k^{2} + 16$ $(f(k))^2 = k^2 + 16$
$20 = k^{2} + 16$ $20 = k^2 + 16$
$4 = k^{2}$ $4 = k^2$
$k = 2$ $k = 2$ (since $k > 0$ $k > 0$ ).

Therefore

$f (k + 1) = f (2 + 1) = f (3) = \sqrt{3^{2} + 16} = \sqrt{25} = 5$ $f(k + 1) = f(2 + 1) = f(3) = sqrt(3^2 + 16) = sqrt(25) =5$

Hopefully this helps!