Does the series from n=1 to infinity of n*(1/2)^n converge or diverge, and why?
Use the ratio test (plug in n+1 for n, multiply by the reciprocal of the original series, and take the limit of the absolute value of this as x approaches infinity). lim x-->inf (n+1)/2^(n+1) * 2^n/n Combine similar terms and simplify: lim x-->inf (2^n)/(2^(n+1)) * (n+1)/n lim x-->inf (1/2)*(n+1)/n = 1/2 Since the limit equals 1/2, which is less than 1, the series converges by the ratio test.
Find the derivative of f(x) = 5/(e^4x + x^3).
First, since there are no variables in the numerator, we can rewrite the function as f(x) = 5(e^4x + x^3)^(-1). Now we can use the chain rule instead of the quotient rule. For the chain rule, multiply by the exponent, subtract one from the exponent, and then multiply by the derivative of the inside: f'(x) = -5*(e^4x + x^3)^(-2)*(4e^4x + 3x^2). We can rewrite this as f'(x) = -5(4e^4x + 3x^2) / (e^x + x^3)^2.
Write the following expression as a single logarithm: 2log(x) - 3log(y) + (1/2)log(z)
First, write the coefficients as exponents (1/2 power is the square root): log(x^2) - log(y^3) + log(sqrt(z)) Now we can combine the terms into a single log (add-->multiply, subtract-->divide): log(x^2*sqrt(z) / y^3)