Solve f^prime(x)=-3x^2+4x-3u

$$f ^ { \prime } ( x ) = - 3 x ^ { 2 } + 4 x - 3 u$$

Solve for u

$$u=-\frac{x\left(3x-4\right)}{3}$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$-3x^{2}+4x-3u=\frac{\mathrm{d}}{\mathrm{d}x}(f)x$$

Add 3x^{2} to both sides.

$$4x-3u=\frac{\mathrm{d}}{\mathrm{d}x}(f)x+3x^{2}$$

Subtract 4x from both sides.

$$-3u=\frac{\mathrm{d}}{\mathrm{d}x}(f)x+3x^{2}-4x$$

The equation is in standard form.

$$-3u=3x^{2}-4x$$

Divide both sides by -3.

$$\frac{-3u}{-3}=\frac{x\left(3x-4\right)}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$u=\frac{x\left(3x-4\right)}{-3}$$

Divide x\left(-4+3x\right) by -3.

$$u=-\frac{x\left(3x-4\right)}{3}$$

Solve for f

$$f\in \mathrm{R}$$

$$u=-\frac{x\left(3x-4\right)}{3}$$