y=sinx-tanx

Here's my attempt at the derivatives

y ' = cosx-(sec(x))^2

y '' = -sinx-2(secx)*secxtanx

How can I find the points of inflection from that second derivative (where it is zero?)?

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- Apr 15th 2008, 04:32 PMtheowneFind points of inflection of trig function
y=sinx-tanx

Here's my attempt at the derivatives

y ' = cosx-(sec(x))^2

y '' = -sinx-2(secx)*secxtanx

How can I find the points of inflection from that second derivative (where it is zero?)? - Apr 15th 2008, 05:35 PMMathstud28Ok
Check your work against mine...$\displaystyle y=sin(x)-tan(x)$...so $\displaystyle y'=cos(x)-sec(x)^2$ and [/tex]-sin(x)-2sec(x)^2tan(x)[/tex]....you cant find the inflection points without an interval but I will assume it is $\displaystyle [0,2\pi]$...what I would do is graph the curve and find the points of intersection or do this factor out a $\displaystyle sec(x)^2$ to get $\displaystyle -sec(x)^2[sin(x)^3+2tan(x)]=0$ then go $\displaystyle -sec(x)^2[sin(x)[sin(x)^2+csc(x)]]$ and work with that

- Apr 15th 2008, 05:38 PMMathstud28Oh
And I almost forgot...an inflection point occurs where the second derivative is zero or undefined...but set up test intervals based on where the second derivative is undefined, zero, and where the original function is undefined...now test an element of each in the second derivative...where there is a sign change between intervals there is an inflection point