Let `g(x)=f(x)+f(1-x)` and `f''(x)>0AAx in (0,1)dot` Find the intervals of increase and decrease of `g(x)dot`

Let g(x)=f(x)+f(1-x) and f''(x)<0 , when x in (0,1) . Then f(x) is

Let g(x)=f(logx)+f(2-logx)a n df''(x)<0AAx in (0,3)dot Then find the interval in which g(x) increases.

Let g(x)=f(logx)+f(2-logx) and f^(prime prime)(x)<0AAx in (0,3)dot Then find the interval in which g(x) increases.

Let g(x)=2f(x/2)+f(2-x)a n df^('')(x)<0AAx in (0,2)dot Then g(x) increases in (a) (1/2,2) (b) (4/3,2) (c) (0,2) (d) (0,4/3)

Let g(x)=2f(x/2)+f(1-x) and f^(primeprime)(x)<0 in 0<=x<=1 then g(x)

Let g(x) = 2 f(x/2)+ f(2-x) and f''(x) lt 0 forall x in (0,2) . Then calculate the interval in which g(x) is increasing.

Let f'(sinx)lt0andf''(sinx)gt0,AAx in (0,(pi)/(2)) and g(x)=f(sinx)+f(cosx), then find the interval in which g(x) is increasing and decreasing.

Let f(x)=sin^(-1)((2phi(x))/(1+phi^(2)(x))) . Find the interval in which f(x) is increasing or decreasing.

Let f(x)=(sinx)/x and g(x)=|x.f(x)|+||x-2|-1|. Then, in the interval (0,3pi) g(x) is :

Let g(x) be the inverse of f(x) and f'(x)=1/(1+x^(3)) .Find g'(x) in terms of g(x).