
Doug C. answered 09/19/22
Math Tutor with Reputation to make difficult concepts understandable
For part 1:
f(2x+1) = 4x2+4x+8; since (2x+1)2= 4x2+4x+1, looks like we just need to add on a +7.
f(x) = x2+7
f(2x+1) = (2x+1)2+7= 4x2+4x+1+7=4x2+4x+8
For part 2:
f(x) = 5x+6; we want f(/) = 5x2+5x+1, ;let's try putting just x2+x into f(x).
f(x2+x) = 5(x2+x) + 6 = 5x2+5x+6 -- too big by 5, so let's try x2+x-1.
g(x) = x2+x-1
f(g(x)) = 5(x2+x-1) + 6 = 5x2+5x-5+6 = 5x2+5x+1