Solve f^prime(x)=x^2+2ax+1-2a

$$f ^ { \prime } ( x ) = x ^ { 2 } + 2 a x + 1 - 2 a$$

Solve for a

$$a=\frac{x^{2}+1}{2\left(1-x\right)}$$

$$x\neq 1$$

Steps for Solving Linear Equation

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

$$x^{2}+2ax+1-2a=\frac{\mathrm{d}}{\mathrm{d}x}(f)x$$

Subtract x^{2} from both sides.

$$2ax+1-2a=\frac{\mathrm{d}}{\mathrm{d}x}(f)x-x^{2}$$

Subtract 1 from both sides.

$$2ax-2a=\frac{\mathrm{d}}{\mathrm{d}x}(f)x-x^{2}-1$$

Combine all terms containing a.

$$\left(2x-2\right)a=\frac{\mathrm{d}}{\mathrm{d}x}(f)x-x^{2}-1$$

The equation is in standard form.

$$\left(2x-2\right)a=-x^{2}-1$$

Divide both sides by 2x-2.

$$\frac{\left(2x-2\right)a}{2x-2}=\frac{-x^{2}-1}{2x-2}$$

Dividing by 2x-2 undoes the multiplication by 2x-2.

$$a=\frac{-x^{2}-1}{2x-2}$$

Divide -x^{2}-1 by 2x-2.

$$a=-\frac{x^{2}+1}{2\left(x-1\right)}$$

Solve for f

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

$$a=\frac{x^{2}+1}{2\left(1-x\right)}\text{ and }x\neq 1$$