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Determine the function f(x) if f(0)=0 and f'(x)=2x+1.
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Since the result of the first derivative is a linear function, we'll consider f(x) as being a quadratic function.
f(x) = ax^2 + bx + c
We'll consider the constraint from enunciation:
f(0) = 0
We'll substitute x by 0 in the expression of the quadratic:
f(0) = a*x^2 + b*0 + c
f(0) = c
But f(0) = 0, so c = 0.
Now, we'll differentiate f(x):
f'(x) = (ax^2 + bx + c)'
f'(x) = 2ax + b (1)
We'll impose the other constraint given by enunciation:
f'(x) = 2x + 1 (2)
We'll put (1) = (2):
2ax + b = 2x + 1
For the identity to hold, we'll have to impose that the coefficients of x from both sides have to be equal and the terms that do not contain x from both sides, to be equal.
2a = 2
a = 1
b = 1
The expression of the original function is:
f(x) = x^2 + x
Posted by giorgiana1976 on October 8, 2010 at 8:34 PM (Answer #1)
High School Teacher
f'(x) = 2x+1.
So to determine f(x) we have to integrate 2x+1.
f(x) = Int (2x+1)dx
f(x) = Int (2x)dx +Int dx ++constant
f(x) =2(x^2)/2 +x +C = x^2+x+C..........(1)
But given f(0) = 0. So put x= 0 in (1) and we get:
f(0) = 0^2+0+ C= o. So C = 0
Therefore f(x) = x^2+x
Posted by neela on October 8, 2010 at 8:39 PM (Answer #2)
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