A function `g(x)` is defined `g(x)=2f(x^2/2)+f(6-x^2),AA x in R` such `f''(x)> 0, AA x in R`, then m`g(x)` has

Let g'(x)gt 0 and f'(x) lt 0 AA x in R , then

A function f is defined by f(x^(2) ) = x^(3) AA x gt 0 then f(4) equals

If g(x) =2f(2x^3-3x^2)+f(6x^2-4x^3-3) AA x in R and f''(x) gt 0 AA x in R then g(x) is increasing in the interval

The function f (x) satisfy the equation f (1-x)+ 2f (x) =3x AA x in R, then f (0)=

A function f : R -> R^+ satisfies f(x+y)= f(x) f(y) AA x in R If f'(0)=2 then f'(x)=

If f(x)+2f(1-x)=x^2+2AA x in R, then f(x) given as

If f(x)+2f(1-x)=x^2+2AA x in R, then f(x) given as

Let f(x+y)=f(x)+f(y) and f(x)=x^2g(x)AA x,y in R where g(x) is continuous then f'(x) is

Consider the function f(x)=(x^(3)-x)|x^(2)-6x+5|, AA x in R , then f(x) is

Let f : R rarr R and g : R rarr R be two given functions defined as f(x) = 3x^(2) + 2 and g(x) = 3x-1, AA x in R Then find [(gof)(x)] at x = 1.