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#

We have:

#(f(x))^2 = (sqrt(x^2 + 16))^2 = x^2 + 16#

We now can solve for #k#.

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

Therefore

#f(k + 1) = f(2 + 1) = f(3) = sqrt(3^2 + 16) = sqrt(25) =5#

Hopefully this helps!