4. Find f (2), f (3), f (4), and f (5) if f is defined recursively by f (0) = f (1) = 1 and for n = 1, 2, . . . a) f (n + 1) = f (n) − f (n − 1). b) f (n + 1) = f (n)f (n − 1). c) f (n + 1) = f (n)2 + f (n − 1)3. d) f (n + 1) = f (n)/f (n − 1).

a) f(2) = 0; f(3) = f(4) = -1; f(5) = 0

b) f(2)=f(3)=f(4)=f(5) = 1

c) f(2) = 2; f(3) = 5; f(4) = 27; f(5) = 734

d) f(2)=f(3)=f(4)=f(5) = 1

Step-by-step explanation:

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!