A differential function `y=f(x)` satisfies `f"(x)-(f(x))^2+5` and `f(0)=1`. Then the equation of tangent at the point where the curve crosses y-axis is

A differentiable function y= f(x) satisfies f'(x)=(f (x))^2+5 and f (0)= 1 . Then the equation of tangent at the point where the curve crosses y-axis, is

A differentiable function y=f(x) satisfies f'(x)=(f(x))^(2)+5 and f(0)=1. Then the equation of tangent at the point where the curve crosses y-axis,is

Determine a differentiable function y=f(x) which satisfies f'(x)=[f(x)]^(2) and f(0)=-(1)/(2). Find also the equation of the tangent at the point where the curve crosses the y-axis.

Let f :[0,2p] to [-3, 3] be a given function defined at f (x) = cos ""(pi)/(2). The slope of the tangent to the curve y = f^(-1) (x) at the point where the curve crosses the y-axis is:

Let f :[0,2p] to [-3, 3] be a given function defined at f (x) = 3 cos ""(x)/(2). The slope of the tangent to the curve y = f^(-1) (x) at the point where the curve crosses the y-axis is:

Let f :[0,2pi] to [-3, 3] be a given function defined at f (x) = 3cos ""(x)/(2). The slope of the tangent to the curve y = f^(-1) (x) at the point where the curve crosses the y-axis is:

Let f be a real valued differentiable function satisfying f(x+y)=f(x)+f(y)-xy-1,AA x,y in R and f(1)=1. Equation ofthe normal to the graph of f(x) at the point where y=f(x) cuts the y-axis,is

A polynomial function f(x) satisfies the condition f(x+1)=f(x)+2x+1. Find f(x) if f(0)=1 Findalso the equations of the pair of tangents from the origin on the curve y=f(x) and compute the areaenclosed by the curve and the pair of tangents.