How to find the linearization of this function($ x^4+y^3-x^3-y^2+x^2+y-2=0$)?

Question

This past paper question about linearization for my calculus 1 course has been bothering me. I spent 5 days trying to solve it.

Considering the equation: $ x^4+y^3-x^3-y^2+x^2+y-2=0$

We accept and can use without justification the fact that, for each number x, there is a unique number $f(x)$ such that $x^4+f(x)^3-x^3-f(x)^2+x^2+f(x)-2=0$ and that the function $f$ defined in this manner is differentiable everywhere.

a) Find the linearization of $f$ at 1. (Justify your answer.)

b) Use the linearization from a) to propose an approximate value for $f(1.001)$.

Answer 1

Letting $x=1$ gives $$ y^3-y^2+y-1=0 \text{ or }(y-1)(y^2+1)=0. $$ So $y=1$ or $f(1)=1$. Differentiating $$ x^4+f^3(x)-x^3-f^2(x)+x^2+f(x)-2=0 $$ gives $$ 4x^3+3f^2(x)f'(x)-3x^2-2f(x)f'(x)+2x+f'(x)=0. $$ Letting $x=1$, one has $$ f'(1)=-\frac32. $$ Thus $$ f(x)=f(1)+f'(1)(x-1)+O((x-1)^2). $$ Using this, it is easy to obtain $f(1.001)$.