Let f(x) be a quartic function such that f(2)=f'(2)=f''(2)=2. If R(x) is the remainder when f(x) is divided by (x-2)
3.
Evaluate R(3).
If f(x) is quartic, then it can be written as Q(x)*(x-2)^3 + R(x) where Q(x) is linear and R(x) is at most quadratic and is the remainder when dividing by (x-2)^3.
Then f(2) = R(2), f'(2) = R'(2) and f'''(2) = R''(2) -- all of the other terms will have at least one factor of (x-2) which will be zero at this point.
Now, if R is at most quadratic, then R(x) = ax^2 + bx + c, R'(x) = 2ax + b and R''(x) = 2a. So substitute:
2a = 2 so a = 1
2a*2 + b = 2 => 4 + b = 2 => b = -2
a*2^2 + b*2 + c = 2 => 4 - 4 + c = 2 => c = 2
so R(x) = x^2 - 2x + 2
Accordingly, R(3) = 9 - 6 + 2 = 5
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Posted by Paul
on 2025-01-26 13:04:48 |
Solution