How to get tangent of inverse of curve?? Ok so my question is. Let $ f(x)=(1/7)x^3+21x-1.$ and let y=g(x) be the inverse function of f. Determine all points on the graph of the inverse function g so that the tangent line is perpendicular to the straight line $ y=-42x+4. $There are two point $P(x1,y1)$ and $Q(x2,y2)$ where y1

So i put the equation into wolfram and got a very long inverse.
$-(7 2^(1/3) 7^(2/3))/(1+x+sqrt(9605+2 x+x^2))^(1/3)+(7/2)^(1/3) (1+x+sqrt(9605+2 x+x^2))^(1/3) $
now i took the derivative of this 
$(14 2^(1/3) 7^(2/3) + 2^(2/3) 7^(1/3) (1 + x + sqrt[9605 + 2 x + x^2])^(2/3))/(6 Sqrt[9605 + 2 x + x^2] (1 + x + Sqrt[9605 + 2 x + x^2])^(1/3))$ 
i equated the derivative to $1/42$ 
and solved for x and got $x=195$ and $x=-197$ is this correct so far? and how do i solve for the rest?

The curve $y=g(x)$ has equation that can be written in implicit form as
$$\frac{y^3}{7}+21y-1=x.$$
Differentiate. We get 
$$\frac{dy}{dx}\left(\frac{3y^2}{7}+21\right)=1.\tag{1}$$ 
We want the tangent line to $y=g(x)$ to be perpendicular to a line with slope $-42$. So we want the tangent line to $y=g(x)$ to have slope $\frac{1}{42}$.
Set $\frac{dy}{dx}=\frac{1}{42}$ in (1) and solve for $y$. We get after a while $y=\pm 7$. Now we can find the appropriate values of $x$.
Remark: The curve $y=g(x)$ is obtained by reflecting the curve $y=f(x)$ in the line $y=x$. So if a certain tangent line to $y=g(x)$ has slope $\frac{1}{42}$, then after reflection the correspnding tangent line to $y=f(x)$ has slope $42$, the reciprocal. It is easy to find the points $(a,b)$ on $y=f(x)$ where the tangent line has slope $42$. Now reflect back, and we obtain the appropriate points for $y=g(x)$. 
