f'(x) = 2x^2 - 6x + 7 find f(x) if f(0) = -8.

We are given the first derivative of the function and we need to determine the function.

Then we know that:

f(x) = integral of f'(x).

==> f(x) = intg (2x^2 - 6x + 7) dx

==> f(x) = intg (2x^2) dx - intg (6x) dx + intg 7 dx

==> f(x) = 2x^3/3 - 6x^2/2 + 7x + C

==> f(x) = (2/3)x^3 - 3x^2 + 7x + C

But we are given that f(0) = -8.

Then we will substitute with x= 0.

==> f(0) = 0 - 0 + 0 + C = -8

==> C = -8.

Then, we will substitute with c= -8 into the function.

**==> f(x) = (2/3)x^2 - 3x^2 + 7x - 8**

We are given the derivative of f(x) as f'(x) = 2x^2 - 6x + 7. We have to find f(x) if f(0) = -8.

Now Int [2x^2 - 6x + 7]

=> Int [2x^2] - Int [6x] + Int [7] + C

=> 2*x^3/3 - 6x^2/2 + 7x + C

=> (2/3)x^3 - 3x^2 + 7x + C

So f(0) = (2/3)*0^3 - 3*0^2 + 7*0 + C = -8

=> C = -8

Therefore f(x) = (2/3)x^3 - 3x^2 + 7x - 8

**The required function is f(x) = (2/3)x^3 - 3x^2 + 7x - 8.**

If f'(x) = 2x^2 - 6x + 7 find f(x) if f(0) = -8..

We know that f(x) = Int f'(x) dx.

So if f'(x) = 2x^2-6x+7 , then f(x) = Int (2x^2-6x+7).dx

Int (2x^2-6x+7) dx = Int (2x^2-Int(6xdx)+ Int 7dx + C.

Int (2x^2-6x+7) dx = 2*(1/3)x^3 -6*(1/2)x^2+ 7x+C.

Int (2x^2-6x+7) dx = (2/3)x^3-3x^2+7x +C.

Therefore f(x) = (2/3)x^3-3x^2+7x +C.

Put x= 0 in f(x) = (2/3)x^3-3x^2+7x +C and we get:

f(0) = 0+C. But f(0) = -8.

So c= -8. So rewite with c = -8 in f(x) = (2/3)x^3-3x^2+7x +C :

f(x) = (2/3)x^3-3x^2+7x -8.

If f'(x) = 2x^2 - 6x + 7, that means that the function f(x) is a polynomial of 3rd order:

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

If f(0) = -8, we'll substitute x by 0 and we'll get:

f(0) = d

**d = -8 **

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

We'll compare f'(x) with the given f'(x) and we'll get:

3a = 2

**a = 2/3**

2b = -6

**b = -3**

**c = 7**

**The function f(x) = 2x^3/3 - 3x^2 + 7x - 8.**