Find f(x) by solving the initial value problem.

f '(x) = 3ex - 4x; f(0) = 10
f(x) =

f(x) = 3e^x -2x^2 + c

10 = 3e^0 -2(0)^2+ c
10 =3-0+ c
7= c

F(x) = 3e^x -2x^2 + 7

To find f(x) by solving the initial value problem, we will integrate the given differential equation f'(x) = 3ex - 4x.

Let's start by integrating both sides of the equation:
∫ f'(x) dx = ∫ (3ex - 4x) dx

The integral of f'(x) is simply f(x):
f(x) = ∫ (3ex - 4x) dx

To integrate ∫ 3ex dx, we can use the fact that the integral of ex is ex:
f(x) = 3∫ ex dx - 4∫ x dx

Integrating ex gives us ex:
f(x) = 3ex - 4∫ x dx

To integrate ∫ x dx, we use the power rule of integration which states that the integral of x^n dx is (1/(n+1))x^(n+1):
f(x) = 3ex - 4 * (1/2)x^2 + C

Now we have the general solution to the differential equation. We can use the initial condition f(0) = 10 to find the constant C.

Plugging in x = 0 into the equation f(x) = 3ex - 4 * (1/2)x^2 + C:
10 = 3e^0 - 4 * (1/2)0^2 + C

Simplifying, we get:
10 = 3 - 0 + C
10 = 3 + C

Subtracting 3 from both sides, we find:
C = 7

Now we have the specific solution to the initial value problem:
f(x) = 3ex - 4 * (1/2)x^2 + 7