Respuesta :

Answer:

[tex]f(1)=-7[/tex]

Step-by-step explanation:

We are given the derivative:

[tex]f^\prime (x) = 3 x^2 + 2x[/tex]

And the initial condition that f(2)=3.

And we want to find f(1).

So, to find our original function f(x), we will find the antiderivative of f'(x). Hence:

[tex]\displaystyle f(x)=\int f^\prime(x)\, dx=\int3x^2+2x\, dx[/tex]

By integrating, we acquire:

[tex]f(x)=x^3+x^2+C[/tex]

Since we know that f(2)=3:

[tex]3=(2)^3+(2)^2+C[/tex]

It follows that:

[tex]C=-9[/tex]

Therefore, our function is given by:

[tex]f(x)=x^3+x^2-9[/tex]

Therefore:

[tex]f(1)=(1)^3+(1)^2-9 = -7[/tex]

Here we want to solve a differential equation, we will find that our function is f(x) = x^3 + x^2 -9

Solving the differential equation.

So we want to find f(x) such that we know f'(x).

We know that:

f'(x) = 3x^2 + 2x

To get f(x) we just need to integrate, we will get:

[tex]f(x) = \int\limits {3x^2 + 2x} \, dx = x^3 + x^2 + c[/tex]

Where c is a constant of integration.

Now, we know that f(2) = 3, with that we can find the value of c.

f(2) = 2^3 + 2^2 + c = 3

          8 +  4 + c = 3

          c = 3 - 12 = -9

So the function is:

f(x) = x^3 + x^2 -9

If you want to learn more about differential equations, you can read:

https://brainly.com/question/18760518