# Determine the function f(x) if f(0)=0 and f'(x)=2x+1.

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### 3 Answers

f'(x) = 2x+1

We know that:

f(x) = integral f'(x)

= intg (2x+1) dx

= intg 2x dx + intg 1 dx

= 2x^2/2 + x + C

f(x) = x^2 + x + c

But we know that:

f(0) = 0 ==> c = 0

**==> f(x) = x^2 + x**

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

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**

and

**b = 1**

**The expression of the original function is:**

**f(x) = x^2 + x**