The function f(x) is defined by `f(x)={log_[4x-3](x^2-2x+5) if `3/41,,, 4 when x=1} `

The function f(x) is defined by f(x)={log_[4x-3](x^2-2x+5) if 3/4 1 4 when x=1}

If the function f(x) defined by f(x)=(log(1+3x)-log(1-2x))/(x),x!=0 and k

f(x)=log_(3)(5+4x-x^(2))

The function f is defined as f(x) = -4x^(3) - 4x^(2) . What is f(-4) ?

The function f:R rarr R defined as f(x)=(3x^2+3x-4)/(3+3x-4x^2) is :

If the function f:[2,oo)rarr[-1,oo) is defined by f(x)=x^(2)-4x+3 then f^(-1)(x)=

The function f:R rarr R defined as f(x)=(3x^(2)+3x-4)/(3+3x-4x^(2)) is :

If the function f(x) defined as : f(x)=(x^(4)-64x)/(sqrt(x^(2)+9)-5), for x!=4 and =3, for x=4 show that f(x) has a removable discontinuity at x=4

The function f: R ⇒ R is defined by f(x)=cos^2x+sin^4x , r a n g e o f f(x)= (a) [3//4,1) (b) (3//4,1] (c) [3//4,1] (d) ( 3//4,1)