If $g(x) = f(\tan^2x-2\tan x+4)$, $0<x<π/2$, then $g(x)$ is increasing in what interval?

Question

If $f''(x)>0 \;\forall x \in\Bbb R$, $f'(3) = 0$ and $g(x) = f(\tan^2x - 2\tan x + 4),\enspace 0<x<\frac{π}{2} $, then $g(x)$ is increasing in which 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 $x = \frac{π}{4}$ and it gives me $g(\frac{π}{4}) = 0$ but I am stuck here and don't know what to do after this.

Answer 1

$f''(x)>0$ so $f'$ increasing. Therefore: $x<3 \iff f'(x)<f'(3) \iff f'(x)<0 \forall x<3$

$g'(x)= f'(tan^2x-2tanx+4)*(2tanx-2)/cos^2x$

$g'(x)=0 \iff tanx=1 \iff x=\pi/4$

Therefore for $x\in (0,\pi/4): 2tanx-2<0 \iff g'(x)>0$ so g increasing on $(0,\pi/4)$ for $x\in (\pi/4,\pi/2): 2tanx-2>0 \iff g'(x)<0$ so g decreasing on $(\pi/4,\pi/2)$

Answer 2

I see one thing that since tan(x) is increasing on $(0, \dfrac{\pi}{2})$ and $ \tan ( \pi /4)=1$ Therefore the thing inside f in the definition of g’ is increasing on $(\pi /4 ,\pi/2)$ (infact by chain rule $$g’=f’( (\tan x -1)^2+3) \cdot {\sec(x)}^2( \tan x -1) $$ and you can use the condition on f’’