To determine △3[(1 + x)(1 − 3x)(1 + 5x)] where the common interval length is 1, you need to use the finite difference method.
Firstly, expand (1 + x)(1 − 3x)(1 + 5x) to get:
(1 + x)(1 − 3x)(1 + 5x) = 15x³ - 23x² - 3x + 1
Then, calculate the first three forward differences by subtracting successive values of the function:
△f(x) = f(x + 1) - f(x)
△¹f(x) = f(x + 1) - f(x) = (15(x + 1)³ - 23(x + 1)² - 3(x + 1) + 1) - (15x³ - 23x² - 3x + 1)
= 45x² - 73x - 25
△²f(x) = △¹f(x + 1) - △¹f(x) = (45(x + 2)² - 73(x + 2) - 25) - (45(x + 1)² - 73(x + 1) - 25)
= 90x - 146
△³f(x) = △²f(x + 1) - △²f(x) = (90(x + 3) - 146) - (90(x + 2) - 146)
= 90
Finally, the value of △³[(1 + x)(1 − 3x)(1 + 5x)] at x = 0 is equal to the third forward difference divided by the product of the common interval length raised to the power of 3:
△³[(1 + x)(1 − 3x)(1 + 5x)]|x=0 = △³f(x)/(1^3) = 90/1 = 90
Therefore, △³[(1 + x)(1 − 3x)(1 + 5x)] where the common interval length is 1 and x = 0 is equal to 90.