Consider this function in explicit form.

f(n)=3n-4; n>1

Select the equivalent recursively defined function.

A. f(1)= -1; f(n)=f(n-1)+3; n>2

B. f(1)= -1; f(n)=3f(n-1)+3; n>2

C. f(0)= -4; f(n)=3f(n-1)+3; n>2

D. f(0)= -4; f(n)=f(n-1)+3; n>2
asked Nov 8, 2016 in Algebra 1 Answers by marieh_ (220 points)

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1 Answer

f(n)=3n-4 so f(0)=-4, f(1)=-1, f(2)=2, f(3)=5. (f(4)=8, etc., so it's clear that f(n)=f(n-1)+3.)

In C, D f(0)=-4 and in A, B f(1)=-1. So they are consistent.

In A, f(2)=f(1)+3=-1+3=2 the same as f(2) originally and f(3)=f(2)+3=2+3=5 the same as f(3) originally.

The other options (B and C) are not consistent with the original f(n). So the answer is A. D appears to be the same as A because f(0)=-4 and f(1)=-1 for both A and D.

answered Nov 8, 2016 by Rod Top Rated User (487,620 points)
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