f(x) = 3x/(1+x²)
f'(x) = [(3)(1+x²) - (3x)(2x)]/(1+x²)²
= (3+3x²-6x²)/(1+x²)² = (3-3x²)/(1+x²)²
f(4) = 12/17 = 0.70588
f'(4) = -9/289 = -0.03114
So, we want a line through (4/0.70588) with slope -0.03114
(y-0.70588)/(x-4) = -0.03114
y = -0.3114x + 0.83044
If f(x)=(3x)/(1+x^2)
find f′(4).
Use this to find the equation of the tangent line to the curve y=3x1+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:
1 answer
1 answer