Given the derivative of f(x) is f'(x) = 3x^2 - 5x + 1.

We know that:

f(x) = integral f'(x)

==> f(x) = intg (3x^2 - 5x + 1) dx

= 3x^3/3 - 5x^2/2 + x + C

==> f(x) = x^3 - (5/2)x^2 + x + C

But we are given that : f(0) = 4

Then we will substitute with x= 0

==> f(0) = 0 - 0 + 0 + C = 4

Then we conclude that C = 4

**==> f(x) = x^3 - (5/2)x^2 + x + 4**

Since the first derivative is a quadratic, then the original function is a polynomial of 3rd order.

We'll put f(x):

f(x) = ax^3 + bx^2 + cx + d

To determine f(x), we'll have to calculate the coefficients a,b,c,d.

We know, from enunciation, that f(0) = 4.

f(0) = d => **d = 4**

Now, we'll differentiate f(x):

f'(x) = 3ax^2 + 2bx + c

From enunciation, we know that: f'(x) = 3x^2 - 5x + 1

We'll put:

3ax^2 + 2bx + c = 3x^2 - 5x + 1

We'll put the correspondent coefficients as equal:

3a = 3

**a = 1**

2b = -5

**b = -5/2**

**c = 1**

The original function is:

**f(x) = x^3 - 5x^2/2 + x + 4**

To find f(x) if f'(x) = 3x^2 - 5x + 1 if f(0) = 4.

If f'(x) = 3x^2 - 5x + 1 , then f(x) = Integral f'(x) dx.

Integral f'(x) dx = Integral {3x^2 - 5x + 1 }

Integral f'(x) dx = Integra (3x^2)dx - Integral (5x) dx +Integral dx

Integral f'(x) dx = 3* (1/3)x^3 - 5 (1/2) x^2 +x+ C.

Integral f'(x) dx = x^3-2.5x^2+x+C.

Therefore f(x) = x^3-2.5x^2+x+C.

Therefore f(0) = x^3-2.5*0^2+0 +C = 4 as f(0) = 4 given.

Therefore C = 4.

So f(x) = x^3-5x^2 +x +4.