Let `f(x)=x^3-3x^2+6AAx in \mathbb{R}` and `g(x)={m a x :f(t ; x+1lt=tlt=x+2,-3lt=xlt=0 1-x ,xgeq`

Let f(x) = x^(3) - x^(2) + x + 1 and g(x) = {{:(max f(t)",", 0 le t le x,"for",0 le x le 1),(3-x",",1 lt x le 2,,):} Then, g(x) in [0, 2] is

Let f(x)={x+1,x >0 \n 2-x ,xlt=0 and g(x)={x+3,x 0)g(f(x)).

Let f(x)={x^2-4x+3,x<3 and x-4,xgeq3 }a n d g(x)={1,x<1 and x^2-x,1lt=x<4 and x-3,xgeq4 then function f(x)dotg(x) is discontinuous at (a)exactly 1 point (b) exactly 2 points (c)exactly 3 points (d) None of these

Let f(x)={x+1,x >0, 2-x ,xlt=0 and g(x)={x+3,x 0)g(f(x)).

If f(x)=max{x^3, x^2,1/(64)}AAx in [0,oo),t h e n f(x)={x^2,0lt=xlt=1x^3,x > 0 f(x) = { 1/(64), 0 lt= x lt = 1/4 x^2, 1/4 1 f(x)={1/(64),0lt=xlt=1/8x^2,1/8 1 f(x)={1/(64),0lt=xlt=1/8x^3,x >1/8

Let f(x)=x^2-2x ,x in R ,a n dg(x)=f(f(x)-1)+f(5-f(x))dot Show that g(x)geq0AAx in Rdot

Let f(x)=(x^(2)-2x+1)/(x+3),f i n d x: (i) f(x) gt 0 (ii) f(x) lt 0

Let f(x)={(x^(2)-4x+3",",x lt 3),(x-4",",x ge 3):} and g(x)={(x-3",",x lt 4),(x^(2)+2x+2",",x ge 4):} . Describe the function f//g and find its domain.

Let f(x)={(x^(2)-4x+3",",x lt 3),(x-4",",x ge 3):} and g(x)={(x-3",",x lt 4),(x^(2)+2x+2",",x ge 4):} . Describe the function f//g and find its domain.

If f(x) = 2x^(3)-15x^(2)+24x and g(x) = int_(0)^(x)f(t) dt + int_(0)^(5-x) f(t) dt (0 lt x lt 5) . Find the number of integers for which g(x) is increasing.