The curvature of the function `f(x)=x^(2)+2x+1` at `x=0` is
(a) `(2)/(5^("3/2"))`
(b) `(2)/(5^("1/2"))`
(c) `2/5`
(d) `(2)/(5^("-3/2"))`

Given the function f(x) = x^(2) + 2x + 5 . Find f(-1), f(2), f(5).

Find the range of the function f(x)=cot^(-1)log_(0.5)(x^(4)-2x^(2)+3)

The function f(x)=sum_(k=1)^(5)(x-K)^(2) assumes then minimum value of x given by (a) 5 (b) (5)/(2) (c) 3(d)2

The function f(x)=sum_(r=1)^(5)(x-r)^(2) assuming minimum value at x=(a)5(b)(5)/(2)(c)3(d)2

The value of f(0), so that the function f(x)=((27-2x)^((1)/(3))-3)/(9-3(243+5x)^(1/5))(x!=0) is continuous,is given by (a) (2)/(3)(b)6(c)2(d)4

(2/3)^(-5) is equal to ((-2)/3)^5 (b) (3/2)^5 (c) (2x-5)/3 (d) 2/(3x5)

if f(x)=2x^(2)+3x-5 , the what is f'(0)+3f'(-1)=

A fraction equivalent to 2/3\ is (2+3)/(3+3) (b) (2-1)/(3-1) (c) (2\ x\ 5)/(3\ x\ 5) (d) (2+5)/(3+5)

Let f:A rarr B be an invertible function.If f(x)=2x^(3)+3x^(2)+x-1, then f^(-1)(5)=

Let f (x)=cos(pi/x) and D_+={x: f (x)>0} . Then D_+ contains (a) (2/5,1/2) (b) (1/2,2/3) (c) (2/3,-1/2) (d) (-pi,-1/2)