consider the function
*Find the equation of the tangent to the curve when x=2

Question

consider the function 
*Find the equation of the tangent to the curve when x=2

anonymous · Answer

$$f(x)=(x+b)^2*(2x-1)$$

anonymous · Answer

above equation is the subjected function of the queation..! help me!

hartnn · Answer

do u know how to find slope of tangent from f(x) ??

hartnn · Answer

NO?

hartnn · Answer

slope of tangents at x=2 is given by derivative of f(x) at x=2
does this ring the bell ?

hartnn · Answer

tell me what u get f'(x) as..... ?

anonymous · Answer

its $$f'(x)=6x^2+8xb+2b^2-2x-2b$$

hartnn · Answer

the slope of  tangent at x= 2 is given by f'(2)
so just put x=2 there, and find slope.

anonymous · Answer

so $$y-f(x)=f'(x)(x-2)$$ seems to be right?

hartnn · Answer

the point will be (2,f(2))

hartnn · Answer

$y-f(2)=f'(2)(x-2)$

hartnn · Answer

and slope = f'(2)

hartnn · Answer

ok?

anonymous · Answer

thank you it seems much easirer!

hartnn · Answer

yes, its easy :) 
welcome ^_^

anonymous · Answer

then how can i find the value of b in exact form if the straight line with equation y=4x-2 intersects y=f(x) at the minimum turning point? what would be the easiest step?