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
Explanation:
We have:
(f(x))2=(√x2+16)2=x2+16
We now can solve for
(f(k))2=k2+16
20=k2+16
4=k2
k=2 (sincek>0 ).
Therefore
f(k+1)=f(2+1)=f(3)=√32+16=√25=5
Hopefully this helps!