Question

If $g\left(x\right)=f\left(x\right)+f(2a-x)$ and $f^{\prime \prime }\left(x\right)>0$ for all $x\in [0,a]$, then shown that $g\left(x\right)$ is decreasing on $[0,a]$

Answer 1

$g^{\prime }\left(x\right)=f^{\prime }\left(x\right)=f^{\prime }(2a-x)$

Given $\Rightarrow f">0$

$\Rightarrow f^{\prime }$ is increasing function.

Now,

if $x\in [0,a]$

$\Rightarrow 2a-x\in \left[a.2a\right]$

$\forall x_1\in [0,a]$ and $\forall x_2\in [2a-x]$

$\Rightarrow x_2>x_1$

$\Rightarrow f^{\prime }\left(x_2\right)>f^{\prime }\left(x_1\right)$

$\Rightarrow f^{\prime }(2a-x)>f^{\prime }\left(x\right)$

$\Rightarrow f^{\prime }\left(x\right)-f^{\prime }(2a-x)<0$

$\Rightarrow g^{\prime }\left(x\right)<0$

$\Rightarrow g$ is a decreasing function