Question

If the normal to f(x)= (3x/(1+x)) at (2,2) cuts the axes at B and C, determine the legnth of BC.

Answer 1

find the derivative of f(x)
find the value of that derivative at x = 2
call that the slope at (2,2)
then the normal has slope m = -1/derivative
find equation of that normal line, we know m already
y = m x + b
put in (2.2) to get b
Now all you have to do is find the x and y axis intercepts of that straight line.

Answer 2

f(x)= (3x/(1+x))
dy/dx = [(1+x)(3) - 3x ]/(1+x)^2
when x = 2
dy/dx = [9-6]/9 = 3/9 = 1/3
so the slope of the normal m = -3
You take it from there.