What exponential function is the best fit for the data in the table? x f(x) 2 −3 3 0 4 12 f(x) = 4(4)x − 1 + 4 f(x) = 4(4)x − 1 − 4 f(x) = one fourth(4)x − 1 + 4 f(x) = one fourth(4)x − 1 − 4

fourth option



Explanation:

1) the pair x = 3 f(x) = 0, leads you to probe this:

f(3) = 0 = A [4 ^ (3 - 1) ] + C = 0

=> A [4^2] = - C

A[16] = - C

if A = 1/4

16 / 4 = 4 => C = - 4

That leads you to the function f(x) = [1/4] 4 ^( x - 1) - 4

2) Now you verify the images for that function for all the x-values of the table:

x = 2 => f(2) + [1/4] 4 ^ (2 - 1) - 4 = [1/4] 4 - 4 = 4 / 4 - 4 = 1 - 4 = - 3 => check

x = 3 => f(3) = [1/4] 4^ (3 - 1) - 4 = [1/4] 4^2 - 4 = 16 / 4 - 4 = 4 - 4 = 0 => check

x = 4 -> f(4) = [1/4] 4^ (4-1) - 4 = [1/4] 4^(3) - 4 = (4^3) / 4 - 4 = 4^2 - 4 = 16 - 4 = 12 => check.

Therefore, you have proved that the answer is the fourth option.

The density of the cube is 40