# Finding the Zeroes of a Second Derivative to Determine Points of Inflection

I have been told to graph the function $y=\cos \left(x^2\right)$ for $-2π ≤ x ≤ 2π$. I have determined key features of the graph but need help when it comes to determining the points of inflection for the graph.

I have determined the second derivative of the function $y=\cos \left(x^2\right)$ to be $y''=-2\sin \left(x^2\right)-4x^2\cos \left(x^2\right)$. When I equal the second derivative to 0 and try to solve I get to $0=\frac{\tan \left(x^2\right)}{-2x^2}$ and am stuck after that. What should I be doing to solve for possible values of x?

• actually we get $$-2x^2=\tan(x^2)$$ – Dr. Sonnhard Graubner Aug 1 '15 at 21:23
• this equation can only be solved by a numerical method – Dr. Sonnhard Graubner Aug 1 '15 at 21:26
• the trivial solution is $$x=0$$ – Dr. Sonnhard Graubner Aug 1 '15 at 21:29
• How would I go about to solve for the rest of the x-intercepts? – funmath Aug 1 '15 at 21:31
• try the Newton Raphson method – Dr. Sonnhard Graubner Aug 1 '15 at 21:37

You properly wrote that $$y''=-2\sin \left(x^2\right)-4x^2\cos \left(x^2\right)$$ and you look for the zero's of this expression in the range $-2\pi\leq x\leq 2 \pi$. First of all, you can restrict the range to $0\leq x\leq 2 \pi$ since if $x$ is a solution, $-x$ is another one.

There are many solutions beside the trivial $x=0$. You cannot obtain any closed form solution and only numerical methods (such as Newton, providing good starting values) will provide them.

If I may suggest, do not change the problem to $\tan(x^2)+2x^2=0$ because of the discontinuities of the tangent. As you wrote it, $y''$ is a nice, smooth and continuous function and, providing good guesses as starting values, you should get all solutions without any trouble.

Looking at the plot of $y''$ shows roots around $1.4$, $2.2$, $2.8$, $3.3$, $3.8$, $4.2$, $4.5$, $5.2$, $5.5$, $5.8$, $6.3$. Starting from any of this guesses $x_0$ and applying Newton update will give $$x_{n+1}=x_n-\frac{\sin \left(x_n^2\right)+2 x_n^2 \cos \left(x_n^2\right)}{6 x_n \cos \left(x_n^2\right)-4 x_n^3 \sin \left(x_n^2\right)}$$ For illustration purposes, let us apply the method for the fifth root using $x_0=3.8$; so, the following iterates are generated $$x_1=3.764594249$$ $$x_2=3.764629076$$ $$x_3=3.764629075$$ which is the solution for ten significant figures.

Edit

For sure, you could make life easier setting $x^2=z$ and solve $$\sin(z)+2z\cos(z)=0$$ instead. In this case, the update formula should be $$z_{n+1}=z_n+\frac{\sin (z_n)+2 z_n \cos (z_n)}{2 z_n \sin (z_n)-3 \cos (z_n)}$$

If you look at the equation in $z$ as the ntersection of curves $\tan(z)$ and $-2z$, you notice that they will intersect just above $z=(2k+1)\frac \pi 2$ that is to say $x_0=\sqrt{(2k+1)\frac \pi 2}$. Performing one Newton iteration will give $$x_1=\sqrt{(2k+1)\frac \pi 2}+\frac{1}{\sqrt{2} \pi ^{3/2} (2 k+1)^{3/2}}$$ This will then give you improved estimates.

The solution of the equation 0=tan(x^2)/2x^2 is sqrt(k(pie)) where k is -2, -1, 0, 1 and 2. The equation actually becomes -2x^2=tan(x^2). So the right solution to your trigonometric equation is 0.