Question

An exponential function is given by the equation y=3x. Using the points xx and x+1x+1, show that the y-values increase by a factor of 3 between any two points separated by x2−x1=1.(4 points)

Answer 1

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:

$\frac{f(x + 1)}{f(x)} = \frac{3^{x + 1}}{3^x}$

Here we can use the properties:

$a^n*a^m = a^{n + m}$

$\frac{a^n}{a^m} = a^{n - m}$

Using these in the quotient equation we get:

$\frac{f(x + 1)}{f(x)} = \frac{3^{x + 1}}{3^x} = \frac{3^x*3^1}{3^x} = \frac{3^x}{3^x}*3 = 1*3 = 3$

Then:

$\frac{f(x + 1)}{f(x)} = 3$

$f(x + 1) = 3*f(x)$

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.