Find the equation of the line tangent to the function f(x)= square root of (x^3 + 4x) through the point (2,4)

Answer 1

f(x) = √(x^3+4x)
f'(x) = (3x^2+4) / 2√(x^3+4x)
f'(2) = 16/8 = 2

So, now you have a point and a slope. The line is

y-4 = 2(x-2)
y = 2x

See the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3D%E2%88%9A%28x^3%2B4x%29%2C+y%3D2x+for+x%3D-1..3