# Trig Derivatives (Curve Sketching)

#### pahuja

Have to sketch a graph of y=cos(x^2) from -2π≤x≤2π using curve sketching methods. I have found intercepts and critical points and am now finding the inflection points but am stuck as to how to solve for "x" in second derivative, the following is what I have done so far:
y"=-2(sin(x^2)+2x^2cos(x^2))
0=-2(sin(x^2)+2x^2cos(x^2))
0/-2=sin(x^2)+2x^2cos(x^2)
-2x^2cos(x^2)=sin(x^2)
-2x^2=sin(x^2)/cos(x^2)
-2x^2=tan(x^2)

If someone could show the next steps to find inflection points from -2π≤x≤2π that would be appreciated.

#### DarnItJimImAnEngineer

The left-hand side will always be negative. Therefore the inflection points must be where the tangent is negative. One each in the ranges $-\frac{3\pi}{2} < x^2 < -\pi$, $-\frac{\pi}{2} < x^2 < 0$, $\frac{\pi}{2} < x^2 < \pi$, and $\frac{3\pi}{2} < x^2 < 2\pi$.

Since you're graphing, you only need approximate values of $x$. Might I suggest the Newton-Raphson method?
Start with a guess for $x$ in the desired range, then update the guess using
$\displaystyle x_{new} = x_{previous} - \frac{y''(x_{previous})}{y'''(x_{previous})}$.
It should only take a couple iterations to get a good approximation for $x$ for each range.

idontknow
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