F(x)=320(1/4)^x over the interval 1<x<3?

Answer 1

F(1) = 320(1/4)^1 = 320(1/4)= 80

F(3) = 320(1/4)^3 = 320/64 = 5

F(2) = 320(1/4)^2 = 320/16 = 20

As x goes larger, F(x) decreases. the x-axis is an asymptote, as x goes to infinity, F(x) approaches zero

F(x) = 320(1/4)^x = 320(4)^-x = 320/4^x

Answer 2

Hi James,

Not quite sure what your question is here, but if you are asking what the range of f(x) is over the interval 1 < x < 3, then here's how I would solve it:

Substitute 1 for x into the equation and solve for f(1).

Then substitute 3 for x into the equation and solve for f(3).

f(1) = 320 x (1/4)¹ = 80

f(3) = 320 x (1/4)³ = 5

Therefore, the range of f(x) over the interval 1 < x < 3

is 5 < y < 80