Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, find n(A ∩ B).

Using the formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B). then n(A ∩ B) = n(A) + n(B) - n(A ∪ B) = 20 + 28 - 36 = 48 - 36 = 12

Newton's Method applied to a cubic equation 3. Let f(x) = 4 ln(x) - x. a. Differentiate f and find any critical points. Determine the domain of f and sketch the graph. b. Use Newton's Method to approximate the value of the x-intercept. Start with x0 = 1 and perform two iterations to give the approximation of one zero. Then let x0 = 8 and find x1 to approximate the other zero.

a. The derivative of f is f '(x) = 4/x - 1. The only critical point is when 4/x - 1 = 0, so x = 4. It follows that there is a maximum at (4, 4 ln(4) - 4) = (4, 1.5452). The domain is x > 0. b. The Newton's formula is given by It follows that when x0 = 1, The actual intercept is x = 1.4296. The other intercept is found by starting at x0 = 8. The first Newton iterate is The actual intercept is x = 8.6132.

Find the derivative of f(x)\, with respect to x\,: f(x)=\ln(7x^{2}e^{x}\sin x)\,

At first glance, this problem looks to be very complicated but all we need are the laws of logarithms which allow us to split up products inside logarithms and turn them into sums of logarithms. f(x)=\ln(7x^{2}e^{x}\sin x)=\ln 7+2\ln x+x\ln e+\ln(\sin x)\, Now, we can take the derivatives of the seperate parts. f'(x)={\frac {2}{x}}+1+{\frac {\cos x}{\sin x}}={\frac {2}{x}}+1+\cot x\,