# If $y=f(x)$ is a function derivative implicitly for the locus $y^3+2xy^2+x=4$, how I find $f(1)$??

Question: Suppose that $y = f(x)$ is a derivative function given implicitly from the equation $y^3+2xy^2+x=4$. Suppose that, also, $1\in Df$

a) Compute $f(1)$.

b) Determine the equation of the tangent line to $f$ in $x=1$.

I don't know How to find $f(1)$. I know that $F(x,y)= 0$ defines implicitly a function derivative $y= f(x)$. But when I differentiate I have this:

\begin{gather} \frac{d}{dx}[y^3+2xy^2+x] = \frac{d}{dx} \\ 3y^2\frac{dy}{dx}+2y^2+4xy\frac{dy}{dx}+1 =0 \\ 3y^2\frac{dy}{dx}+4xy\frac{dy}{dx} =-1-2y^2 \\ \frac{dy}{dx}=\frac{-1-2y^2}{3y^2+4xy} \end{gather}

Where do I deduce $f(1)$ from?

Because I need to determine the equation of the line, but I can't without this result.

• What do you need derivatives for? If $y=f(x)$ and $y^3 + 2xy^2 + x=4$ then just plug in: $y^3 + 2y^2 - 3 = 0$. Where a solution for $y$ is a solution of $f(1)$. Youre going to have three possible answers. You dont have any other information to work with. – CogitoErgoCogitoSum Jun 3 '18 at 19:10
• you have a typo in your answer. $\frac{dy}{dx}=\frac{-1-2y^2}{3y^2+4xy}$ – tien lee Jun 3 '18 at 19:11
• I'm reading in between the lines. @CogitoErgoCogitoSum Sometimes the foreign speakers can't communicate well enough. Look at the last sentence of the explanation: He needs the derivative for the equation of the (tangent) line. – Christopher Marley Jun 3 '18 at 19:15
• @CogitoErgoCogitoSum I reported you to the moderator. In the meanwhile, you can read Be nice. – user539887 Jun 3 '18 at 19:30
• I left “derivative function”, but it probably is “explicit function”. – egreg Jun 3 '18 at 20:10

You can find $f(1)$ from the original equation because it means finding the $y$-value when $x=1$.
$$y^3+2xy^2+x=4$$
Substitute in $x=1$ and you get
$$\begin{split} 0 &= y^3+2y^2-3\\ &=(y-1)(y^2+3y+3) \end{split}$$
Note that $\sqrt{3^2-4(3)}$ is complex so $y=1$ is the only choice.