33Hey thats not the fractioanl part na {f(x)}2
If not then whats the problem
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!!!!!!!!!
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5 Answers
33
1if 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@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
6thats again! i was taking it to be a greatest integer function! thanks rohan 2007!!!!!!!!!!!