T and N 14

determine a differentiable function y=f(x) which satisfies f'(x)={f(x)}2 and f(0)=-1/2. Find also the equation to the tangent at the point where the curve crosses the y-axis!!!!!!!!!

5 Answers

33
Abhishek Priyam ·

Hey thats not the fractioanl part na {f(x)}2

If not then whats the problem

1
Rohan Ghosh ·

if it is not the fractional part

as
f'(x)=f(x)2

d(f(x))/f(x)2 =dx

placaing limits from 0 to x

-[1/f(x)] = x

=>

1/f(x)-1/f(0) = -x

f(x) = -1/(x+2)

f'(x) =f(x)2

and the point of crossing y axis = (0,-1/2)
then you will get the equation of the tangent

1
Philip Calvert ·

@Rohan

I dont think for the slope second part we have to use our own answer
the inf given is enuf
f'(0)=f(0)2=1/4
btw you may have forgotten the bracket

f(x)=-1/(x+2) ;D

1
Rohan Ghosh ·

yup

and yup..

6
Aakash Sharawat ·

thats again! i was taking it to be a greatest integer function! thanks rohan 2007!!!!!!!!!!!

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