Question

Estimate the slope of the tangent line of the function y(t)=\sqrt{(3t+1} at the point t=2.

Answer 1

Answer: 0.567

Step-by-step explanation:

$y(t)=\sqrt{3t+1} \\\\y'(t)=\dfrac{3}{2\sqrt{3t+1} } \\\\y'(2)=\dfrac{3}{2\sqrt{3*2+1} } =\dfrac{3\sqrt{7} }{14} =0,56694670951384084082177480435127\approx{0.567}$