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

1 Answer
May 27, 2018

f(k+1)=5

Explanation:

We have:

(f(x))2=(x2+16)2=x2+16

We now can solve for k.

(f(k))2=k2+16
20=k2+16
4=k2
k=2 (since k>0).

Therefore

f(k+1)=f(2+1)=f(3)=32+16=25=5

Hopefully this helps!