Question

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

Answer 1

\[f(x)=(x+b)^2*(2x-1)\]

Answer 2

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

Answer 3

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

Answer 4

NO?

Answer 5

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

Answer 6

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

Answer 7

its \[f'(x)=6x^2+8xb+2b^2-2x-2b\]

Answer 8

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

Answer 9

so \[y-f(x)=f'(x)(x-2)\] seems to be right?

Answer 10

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

Answer 11

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

Answer 12

and slope = f'(2)

Answer 13

ok?

Answer 14

thank you it seems much easirer!

Answer 15

yes, its easy :)
welcome ^_^

Answer 16

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?