A particle is moving with acceleration a(t)=36t+14. its position at time t=0 is s(0)=7 and its velocity at time t=0 is v(0)=5. What is its position at time t=10? f(x)=3x6 and f(2)=6 and f(2)=1. Find f(x)= and find f(3)= Consider the function f(x)=4x10+5x54x49 Enter an antiderivative of f(x). Do not enter +c as part of your answer. Answer: +c