Answer:
Here we have the function:
y = f(x) = 3^x
Using the values:
x and (x + 1)
We need to find that the y-value increases by a factor of 3.
So we need to prove that:
f(x + 1) = 3*f(x).
Or we can see the quotient:
f(x + 1)/f(x) = 3
Here we can find the values:
f(x + 1) = y = 3^(x + 1)
f(x) = y' = 3^x
If we take the quotient, we get:
[tex]\frac{f(x + 1)}{f(x)} = \frac{3^{x + 1}}{3^x}[/tex]
Here we can use the properties:
[tex]a^n*a^m = a^{n + m}[/tex]
[tex]\frac{a^n}{a^m} = a^{n - m}[/tex]
Using these in the quotient equation we get:
[tex]\frac{f(x + 1)}{f(x)} = \frac{3^{x + 1}}{3^x} = \frac{3^x*3^1}{3^x} = \frac{3^x}{3^x}*3 = 1*3 = 3[/tex]
Then:
[tex]\frac{f(x + 1)}{f(x)} = 3[/tex]
[tex]f(x + 1) = 3*f(x)[/tex]
So we found that the y-value increases by a factor of 3 between any two points x₂ and x₁ such that: x₂ - x₁ = 1.
