If g(x)=f(tan2x−2tan⁡x+4), 0&lt;x&lt;π2, then g(x) is increasing in what interval?I have first differentiated the given equation and found the value of x for which the interior part of f is 3 ,this gives x=π4 and it gives me g(π4)=0 but I am stuck here and don't know what to do after this.

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If g(x)=f(tan2x−2tan⁡x+4), 0&amp;lt;x&amp;lt;π2, then g(x) is increasing in what interval?I have first differentiated the given equation and found the value of x for which the interior part of f is 3 ,this gives x=π4 and it gives me g(π4)=0 but I am stuck here and don&#039;t know what to do after this.

Jason Duke · Accepted Answer

f′′(x)&amp;gt;0 so f′ increasing. Therefore: x&amp;lt;3⇔f′(x)&amp;lt;f′(3)⇔f′(x)&amp;lt;0∀x&amp;lt;3   g′(x)=f′(tan2x−2tan⁡x+4)⋅2tan⁡x−2cos2x   g′(x)=0⇔tan⁡x=1⇔x=π4   Therefore for x∈(0,π4):2tan⁡x−2&amp;lt;0⇔g′(x)&amp;gt;0 so g increasing on (0,π4) for x∈(π4,π2):2tan⁡x−2&amp;gt;0⇔g′(x)&amp;lt;0 so g decreasing on (π4,π2)

If g(x)=f(\tan^2 x−2\tan x+4) , \ 0<x<\frac{\pi}{2}, then g(x) is increasing in what interval

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