[11int(f_(1)-f_(0))/(2f_(1)-f_(0)-f_(2)))times h],[" (a) "(1)/(21)" ."],[" I) FIE2"]

The formula of mode is....a) L+[(f_1-f_0)/(2f_1-f_0-f_2)] xxh b) L-[(f_1-f_0)/(2f_1-f_0-f_2)] xxh c) L+[(f_1-f_0)/(2f_1-f_0+f_2)] xxh d) L+[(f_1-f_0)/(2f_2-f_0-f_2)] xxh

lim_(h rarr 0) (f(2h+2+h^(2))-f(2))/(f(h-h^(2)+1)-f(1)) given that f^(')(2) = 6 , and f^(')(1) =4

If int_(0)^(1) f(t)dt=x^2+int_(0)^(1) t^2f(t)dt , then f'(1/2)is

A derivable function f(x) satisfies the relation f(x)=int_(0)^(1)xf(t)dt+int_(0)^(x)x^(2)f(t)dt. The value of (2f'(1))/(f(1)) is

Let f_(1):R to R, f_(2):[0,oo) to R, f_(3):R to R and f_(4):R to [0,oo) be a defined by f_(1)(x)={{:(,|x|,"if "x lt 0),(,e^(x),"if "x gt 0):}:f_(2)(x)=x^(2),f_(3)(x)={{:(,sin x,"if x"lt 0),(,x,"if "x ge 0):} and f_(4)(x)={{:(,f_(2)(f_(1)(x)),"if "x lt 0),(,f_(2)(f_(1)(f_(1)(x)))-1,"if "x ge 0):} then f_(2) of f_(1) is

Let f_(1):R to R, f_(2):[0,oo) to R, f_(3):R to R and f_(4):R to [0,oo) be a defined by f_(1)(x)={{:(,|x|,"if "x lt 0),(,e^(x),"if "x gt 0):}:f_(2)(x)=x^(2),f_(3)(x)={{:(,sin x,"if x"lt 0),(,x,"if "x ge 0):} and f_(4)(x)={{:(,f_(2)(f_(1)(x)),"if "x lt 0),(,f_(2)(f_(1)(f_(1)(x)))-1,"if "x ge 0):} then f_(2) is

Mode can be calculated by = l+((f_1-f_0)/(2f_1-f_0-f_2))xxh here f_1 represents ......

Let f(1):R to R, f_(2):[0,oo) to R, f_(3):R to R and f_(4):R to [0,oo) be a defined by f_(1)(x)={{:(,|x|,"if "x lt 0),(,e^(x),"if "x gt 0):}:f_(2)(x)=x^(2),f_(3)(x)={{:(,sin x,"if x"lt 0),(,x,"if "x ge 0):} and f_(4)(x)={{:(,f_(2)(f_(1)(x)),"if "x lt 0),(,f_(2)(f_(1)(f_(1)(x)))-1,"if "x ge 0):} Then, f_(4) is

Let f(1):R to R, f_(2):[0,oo) to R, f_(3):R to R and f_(4):R to [0,oo) be a defined by f_(1)(x)={{:(,|x|,"if "x lt 0),(,e^(x),"if "x gt 0):}:f_(2)(x)=x^(2),f_(3)(x)={{:(,sin x,"if x"lt 0),(,x,"if "x ge 0):} and f_(4)(x)={{:(,f_(2)(f_(1)(x)),"if "x lt 0),(,f_(2)(f_(1)(f_(1)(x)))-1,"if "x ge 0):} Then, f_(4) is