In order to determine the intervals on which these functions are increasing, we need to determine the intervals on which their derivatives are positive, since the derivative of a function represents its rate of change and an increasing function has a positive rate of change.
a) y = x^3 + e^(2x). Using the power rule on the first term and the chain rule on the second term, we get y' = 3x^2 + 2e^(2x), which is greater than 0 for all x, since 3x^2 is always greater than or equal to zero and 2e^(2x) is always greater than zero. Therefore, the function is increasing on the inverval (-infinity, infinity).
b) y = 5xe^(-4x). Using the product rule and chain rule, we get y' = 5x*e^(-4x)*(-4) + 5*e^(-4x). Factoring out 5*e^(-4x) from each term, we have y' = 5*e^(-4x)*[-4x+1]. Now 5*e^(-4x) is greater than 0 for all x, since the exponential function is always positive. Thus, y' is greater than 0 whenever -4x+1 is greater than 0. Solving this inequality for x, we get x is less than 1/4. Therefore, the function is increasing on the interval (-infinity, 1/4).
c) y=x^2*e^-(x^x). Using the product rule, power rule, and chain rule, we get y' = x^2*e^-(x^x)*[d/dx(-(x^x))] + 2x*e^-(x^x). In order to find the derivative of -(x^x), we use implicit differentiation. Let w = x^x. Taking ln of both sides, we have lnw = ln(x^x). Using the laws of logarithms on the right side gives lnw = x*lnx. Now, we take the derivative of both sides: d/dx(lnw) = d/dx(x*lnx). Using the chain rule on the left side and the product rule on the right side, we have (1/w)(w') = x(1/x) + 1*lnx. Multiplying both sides by w and simplifying the right side, gives us w' = w(1 + lnx). Finally substituting x^x for w, we have w' = (x^x)(1 + lnx). Thus, the derivative of -(x^x) is -(x^x)(1 + lnx). Plugging this into the expression for y', we have y' = x^2*e^-(x^x)*[-(x^x)(1 + lnx)] + 2x*e^-(x^x). Factoring out x*e^-(x^x) from both terms gives us y' = x*e^-(x^x)*[x*(-(x^x))(1 + lnx) + 2]. Since x^x is not continuous for x less than 0 (and therefore not increasing), it suffices to consider x greater than 0 (x^x is undefined for x = 0). On this interval, x*e^-(x^x) is greater than 0, so y' is positive as long as x*(-(x^x))(1 + lnx) + 2 is greater than zero. Unfortunately, this inequality cannot be solved analytically. The best we can do is use a graphing utility to estimate the interval on which y = x*(-(x^x))(1 + lnx) is positive. Doing so gives us 0<x<(approximately)1.246. Therefore, the original function is increasing on the interval (0,1.246), where the second boundary point is an approximation.
