Answer:
simply take the derivative using euler's identity and the zero's of the bessell function. from there, it's pretty straight forward since it's 9 ohms and 5 ohms which is quite lucky frankly.
when you graph it, you end up with a quadratic of 2.5x^2-39x+5923. assuming you're allowed a ti-89, you want to now make a scatterplot out of that using the list and f5 and all that.
find the LSRL from there.
Explanation:
i have no idea what im talking about
