Please help......
2 answers:
(1)
f(x) = ax^3 + bx^2 + cx + d
In table attached, x and f(x) values are shown
a = 3rd difference/3! = 12/6 = 2
f(0) = d = -4
f(2) = 2(2)^3 + b(2)^2 + c(2) - 4 = -2
12 + 4 b + 2 c = -2
4 b + 2 c = -14 (eq. 1)
f(4) = 2(4)^3 + b(4)^2 + c(4) - 4 = -4
128 + 16 b + 4 c - 4 = -4
16 b + 4 c = -128 (eq. 2)
Multiplying eq. 1 by 2 and then subtracting it to eq. 2 gives
8 b = -100
b = -12.5
Replacing in eq. 1
4(-12.5) + 2 c = -14
2 c = 36
c = 18
The polynomial is 2x^3 - 12.5x^2 + 18x - 4
(2)
My polynomial: x^3 + x^2 + x + 1
In table attached, x and f(x) values are shown
a = 3rd difference/3! = 6/6 = 1
f(0) = d = 1
f(1) = (1)^3 + b(1)^2 + c(1) + 1 = 4
1 + b + c + 1 = 4
b + c = 2 (eq. 1)
f(2) = (2)^3 + b(2)^2 + c(2) + 1 = 15
8 + 4 b + 2 c + 1 = 15
4 b + 2 c = 6 (eq. 2)
Multiplying eq. 1 by 2 and then subtracting it to eq. 2 gives
2 b = 2
b = 1
Replacing in eq. 1
(1) + c = 2
c = 1
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if you don't like to start with a negative. Either one is correct.
Answer:
Step-by-step explanation:
<u>Fundamental Theorem of Calculus</u>
If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.
Given indefinite integral:
If the terms are multiplied by constants, take them outside the integral:
Multiply by the conjugate of 1 - sin(6x) :
Expand:
Simplify:
Learn more about indefinite integration here: