Answer:
Step-by-step explanation:
Given expression is,
f(x) = 8x³ - 6x² - 36x + 27
For zeros of the function,
f(x) = 0
8x³ - 6x² - 36x + 27 = 0
2x²(4x - 3) - 9(4x - 3) = 0
(2x² - 9)(4x - 3) = 0
(2x² - 9) = 0
x² =
x =
x =
Or (4x - 3) = 0
x =
Therefore, all zeros of the given functions are .
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:
Here we can use the properties:
Using these in the quotient equation we get:
Then:
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.
