I will assume you meant
f(x) = 3x/(1 + x^2)
then by the quotient rule
f'(x) = ( 3(1 + x^2) - 3x(2x) )/(1 + x^2)^2
f'(4) = (3(17) - 12(8))/289 = -45/289
so now you have the slope m = -45/289
and your strange point (4, .70588)
Use your grade 9 method of y=mx + b
to find the equation.
If f(x)=(3x)/(1+x2) find f′(4)______.
Use this to find the equation of the tangent line to the curve y=(3x)/(1+x2) at the point (4,0.70588). The equation of this tangent line can be written in the form y=mx+b where m is: ________ and where b is:______
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