Question

Let j(\theta) = 3 \theta^3 + 2j()=3 3 +2. let \theta=1=1, and =0.01. use the formula j(+)j()2 to numerically compute an approximation to the derivative at \theta=1=1. what value do you get? (when \theta = 1=1, the true/exact derivative is \frac{dj(\theta)}{d\theta} = 9 d dj() =9.) 9 9.0003 11 8.9997

Answer 1

Answer:

Ask your question clearly yaarr