Trig Derivatives (Curve Sketching)

Trig Derivatives (Curve Sketching)

Postby Guest » Fri Dec 13, 2019 6:45 pm

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 provide next steps and final answer as to what the values of "x" would be within the -2π≤x≤2π range the would be appreciated.
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Re: Trig Derivatives (Curve Sketching)

Postby HallsofIvy » Thu Jan 02, 2020 3:42 pm

That is an equation that combines "algebraic'' functions ([tex]x^2[/tex]) and "transcendental" functions (sin(x) and cos(x)). There is no analytic method to solve such an equation.

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