Solve f(x+h)-f(x)=-3hx^2+6hx+8h^2x-1h^2+3h^3

$$f ( x + h ) - f ( x ) = - 3 h x ^ { 2 } + 6 h x + 8 h ^ { 2 } x - 1 h ^ { 2 } + 3 h ^ { 3 }$$

Solve for f (complex solution)

$$\left\{\begin{matrix}\\f=-3x^{2}+8hx+6x+3h^{2}-h\text{, }&\text{unconditionally}\\f\in \mathrm{C}\text{, }&h=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply f by x+h.

$$fx+fh-fx=-3hx^{2}+6hx+8h^{2}x-h^{2}+3h^{3}$$

Combine fx and -fx to get 0.

$$fh=-3hx^{2}+6hx+8h^{2}x-h^{2}+3h^{3}$$

The equation is in standard form.

$$hf=-3hx^{2}+8xh^{2}+6hx+3h^{3}-h^{2}$$

Divide both sides by h.

$$\frac{hf}{h}=\frac{h\left(-3x^{2}+8hx+6x+3h^{2}-h\right)}{h}$$

Dividing by h undoes the multiplication by h.

$$f=\frac{h\left(-3x^{2}+8hx+6x+3h^{2}-h\right)}{h}$$

Divide h\left(-3x^{2}+6x+8hx-h+3h^{2}\right) by h.

$$f=-3x^{2}+8hx+6x+3h^{2}-h$$

Solve for f

$$\left\{\begin{matrix}\\f=-3x^{2}+8hx+6x+3h^{2}-h\text{, }&\text{unconditionally}\\f\in \mathrm{R}\text{, }&h=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply f by x+h.

$$fx+fh-fx=-3hx^{2}+6hx+8h^{2}x-h^{2}+3h^{3}$$

Combine fx and -fx to get 0.

$$fh=-3hx^{2}+6hx+8h^{2}x-h^{2}+3h^{3}$$

The equation is in standard form.

$$hf=-3hx^{2}+8xh^{2}+6hx+3h^{3}-h^{2}$$

Divide both sides by h.

$$\frac{hf}{h}=\frac{h\left(-3x^{2}+8hx+6x+3h^{2}-h\right)}{h}$$

Dividing by h undoes the multiplication by h.

$$f=\frac{h\left(-3x^{2}+8hx+6x+3h^{2}-h\right)}{h}$$

Divide h\left(-3x^{2}+6x+8hx-h+3h^{2}\right) by h.

$$f=-3x^{2}+8hx+6x+3h^{2}-h$$

Solve for h (complex solution)

$$h=\frac{\sqrt{100x^{2}-88x+12f+1}}{6}-\frac{4x}{3}+\frac{1}{6}$$

$$h=0$$

$$h=-\frac{\sqrt{100x^{2}-88x+12f+1}}{6}-\frac{4x}{3}+\frac{1}{6}$$

Solve for h

$$\left\{\begin{matrix}\\h=0\text{, }&\text{unconditionally}\\h=-\frac{\sqrt{100x^{2}-88x+12f+1}}{6}-\frac{4x}{3}+\frac{1}{6}\text{; }h=\frac{\sqrt{100x^{2}-88x+12f+1}}{6}-\frac{4x}{3}+\frac{1}{6}\text{, }&f\geq -\frac{25x^{2}}{3}+\frac{22x}{3}-\frac{1}{12}\end{matrix}\right.$$