Use the formula ln(1 + x) = ∞ n = 1 (−1)n − 1xn n = x − x2 2 + x3 3 − x4 4 + for |x| < 1 and your knowledge of alternating series to find the smallest n such that the partial sum sn approximates ln 3 2 to within an error of at most 10−3. confirm this using a calculator to compute both sn and ln 3 2 . (round your answers to six decimal places.)

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation:

b

step-by-step explanation:

step-by-step explanation:

we have

using a graphing tool

see the attached figure

this is a quadratic equation (vertical parabola) open up

the vertex is a > vertex

the domain of the function is all real > interval (-∞,∞)

the range of the function is the > [-6.125,∞)

the function has two x->   and

the function is increasing over the > (0.25,∞)

the function is decreasing over the > (-∞,0.25)

the answer is  

d.

8 m

hope you got it down chief.