# Let $f(x)=x^{4}-6x^{2}+5$. If $P(x_0,y_0)$ is a point such that $y_0>f(x_0)$ and there are exactly two distinct tangents through P

Let $$f(x)=x^{4}-6x^{2}+5$$. If $$P(x_0,y_0)$$ is a point such that $$y_0>f(x_0)$$ and there are exactly two distinct tangents through P drawn to the curve $$y=f(x)$$, then find the maximum possible value of $$y_0$$.

The hint said P is the point of intersection of tangents drawn at inflection points.

I calculated inflection points, and they came out to be $$(1,0), (-1,0)$$.

I calculated tangents at these points and found the point of intersection to be $$(0,8)$$.

So, the answer is $$8$$, which is correct but the doubt is why this is happening this way. What's the importance of inflection point here?

• I suppose there should be another condition that the tangents must be on opposite sides of the $y$-axis. In this case, $a = -1, b = 1$ should give the maximum $y_0$ as $8$. Commented Jun 25, 2021 at 13:53

I don't think the answer is correct.

For example the tangent to the curve at $$x=-\frac32$$ is $$r: y=\frac{9}2x + \frac{53}{16}$$ and the tangent line to the curve at $$x=-\frac14$$ is $$s : y=\frac{47}{16}x + \frac{1373}{256},$$ and they interesect at $$P\left(\frac{21}{16},\frac{295}{32}\right)$$.

EDIT

The correct answer is the following. The supremum of $$y_0$$ is the only real positive $$y$$-solution to the system $$\begin{cases}y&=&x^4-6x^2+5\\ y&=&8x+8,\end{cases}$$ that is the ordinate of the interesction point between the given curve and the tangent line at the inflection point $$(-1,0)$$.

We get $$y_{sup}=32$$ (see dashed line in the figure below).

In order to draw the above conclusion you can use the following observations.

1. Any point $$P$$ whose ordinate is such that $$y> 8|x|+8$$ (that is any point above the the two tangent lines shown in the figure) can be connected to the concave part of the graph $$f(x)=x^4-6x^2+5$$ only with a straight line whose angular coefficient has absolute value greater than $$8$$, which shows that such line cannot be tangent to the curve.
2. On the other hand, consider any point $$P_1$$ on the curve, whose abscissa $$x_{P_1}$$ is, e.g., $$-1, and let $$r_1$$ be the line tangent to the curve at $$P_1$$. By appropriately selecting a second tangency point $$P_2$$ with abscissa $$-3 we can find a second tangent line $$r_2$$ whose intersection with $$r_1$$ is any point $$P\in r_1$$ with $$x_P>x_{P_1}$$.
• So can the problem be solved? Commented Jun 25, 2021 at 15:20
• @Ritam_Dasgupta I have to put some more thought on that. As soon as I can :)
– dfnu
Commented Jun 25, 2021 at 16:15
• @Ritam_Dasgupta Intuitively, $\sup\{y_0\} = \overline y$, where $\overline y$ is the ordinate of the interesection point between the curve and the tangent line in $(-1,0)$. But at the moment do not know how to rigorously prove the assertion.
– dfnu
Commented Jun 25, 2021 at 16:55