calculusderivativesimplicit-differentiationsolution-verificationtangent line
Question: Find the $x$-coordinate of each point where the tangent line of $2y(3−x)=x^4$ is vertical.
My work so far: Using implicit differentiation, we get that $\frac{dy}{dx} = \frac{2x^3 +y}{3-x}.$
Therefore, the derivative is undefined when $3 – x = 0,$ making $x = 3.$
Are there any other solutions to the problem?
If you want a horizontal tangent, then you need to look for where ${dy\over dx}=0$.
Edit: As indicated, solve ${dy\over dx}=0\implies y(2.5-1/x)=0\implies y=0\text{ or }x={2\over 5}$. But $y=0$ is not in the domain of the original relation, $\ln(xy)=2.5x$, so the horizontal tangent occurs only when $x=2/5\implies y=5e/2$.
This is depicted graphically below: the dashed line is the horizontal tangent passing thru $(2/5,5e/2)$ on the graph of $\ln(xy)=2.5x$.
We have that
$$x^3-xy+y^3=0\implies 3x^2dx-dx-xdy+3y^2dy=0 \implies \frac{dy}{dx}=\frac{3x^2-y}{x-3y^2}=0$$
that is
$$3x^2=y \implies (x,y)=(t,3t^2)$$
$$x^3-xy+y^3=0 \iff t^3-3t^3+27t^6=0\iff t^3(27t^3-2)=0$$
that is
$$(x,y)=\left(\frac{\sqrt[3]2}{3},\sqrt[3]4\right)$$
Best Answer
A point on the curve where there is a vertical tangent line is necessarily a point where ${\large{\frac{dy}{dx}}}$ is undefined, so the only points where there might be a vertical tangent line are points on the curve where $x=3$.
But there are no points on the curve where $x=3$ since for $x=3$, the $\text{LHS}$ of the equation is equal to $0$ while the $\text{RHS}$ of the equation is equal to $81$.
Here's another way to understand it . . .
Since there are no points on the curve where $x=3$, the equation can be rewritten as $$ y = \frac{x^4}{2(3-x)} $$ valid for all $x$ in the domain $A=\{x\in\mathbb{R}{\,\mid\,}x\ne 3\}$.
But for $x\in A$, ${\large{\frac{dy}{dx}}}$ exists, hence there are no points on the curve where there is a vertical tangent line.
Note that at $x=3$ there is a vertical asymptote.
A further note . . .
If you have a curve given by an equation of the form $f(x,y)=0$, the points on the curve where there is a vertical tangent line are necessarily points where ${\large{\frac{dy}{dx}}}$ is undefined. But the converse is not true. As an example, for the curve $y=|x|$, at $x=0$, ${\large{\frac{dy}{dx}}}$ is undefined, but there are no points on the curve where there is a vertical tangent line.
In general, for a curve given by an equation of the form $f(x,y)=0$, the points on the curve where there is a vertical tangent line are the points on the curve where ${\large{\frac{dx}{dy}}}=0$.
Thus, the task of finding all such points can be accomplished as follows . . .
$(1)\;\;$Differentiate the equation $f(x,y)=0\;$implicitly with respect to $y$.
$(2)\;\;$Solve for ${\large{\frac{dx}{dy}}}$.
$(3)\;\;$Solve the system $$ \left\lbrace \begin{align*} \frac{dx}{dy}&=\,0\\[4pt] f(x,y)&=\,0\\[4pt] \end{align*} \right. $$
$\qquad$for $x,y$.