Consider the exponential function [tex]f(x) = 57(0.48)^x[/tex].

1. What is the 2-unit growth factor for [tex]f[/tex]?
2. What is the 1-unit growth factor for [tex]f[/tex]?
3. What is the initial value of [tex]f[/tex]?

Question

High School

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Answer :

somyatiwari11 somyatiwari11 · Answer 1 · 2024-05-29T10:30:58-04:00

The 2-unit growth factor for f is 0.2304.

The 1-unit growth factor for f is 0.48.

The initial value of f is 57.

Given an exponential function f(x) = 57(0.48).

Formula for the exponential function is:

f(x) = a*b*x

Here, a= initial value, b = growth factor, x = variable.

The growth factor b can be calculated as 0.48.

In order to calculate the 2-unit growth factor, substitute (x + 2) in the given function. So the function becomes:

f(x+2) = 57(0.48)2

Now the growth factor is calculated using the formula, f(x+2)/f(x).

So the 2-unit growth factor is = 0.48^2 = 0.2304.

Now in order to calculate the 1-unit growth factor, substitute (x + 1) in the given function.

So the function becomes f(x+1) = 57(0.48)

Now the growth factor is calculated using the formula, f(x+1)/f(x).

So the 1-unit growth factor = 0.48.

From the given function, initial value a is 57. Hence the initial value of f is 57.

Therefore, the initial value of f is 57, the growth factors for 2-unit and 1-unit increases are 0.2304 and 0.48 respectively.

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