How to find the equation of the curve? PLEASE HELP, 10 POINTS?

First, you must find a family of equations that satisfy the given differential equation. Thankfully, the give equation is purely in terms of x, meaning we can simply integrate it relative to x to obtain a the family. so the integral of 3/(1+2x)^2 dx = -3/(4x+2) +C (use u substitution to solve, u=2x+1) to find C we must plug in the given initial conditions (1,1/2) for x and y respectively, giving 1/2= -3/(4+2) +C solving for C yields +1. this gives you a final equation y=-3/(4x+2) +1. now, to find the domain on which the gradient (dy/dx) is less than 1/3 we need only plug that into the original equation. dy/dx=3/(1+2x)^2 <1/3 and solving for x gives -2>x and 1<x meaning that the domains on which dy/dx is less than 1/3 are (-infinity,-2) and (1, infinity)

You have 2 questions here: 1.) Given dy/dx and the point (1, 1/2), solve for y(x). 2.) Find all x where dy/dx < 1/3 1.) Integrate and then solve for your integration constant... ∫3/(1+2x)^2 dx = ? u = 1 + 2x du = 2 dx 3/2 * ∫ 1/u^2 du = 3/2 * ∫ u^(-2) du = 3/2 * -1/u + C = -3/2 / (1+2x) +C y(1) = 1/2 = -3/2(1+2(1)) + C 1/2 = -3 / (2 * 3) + C 1/2 = -1/2 + C 1 = C So, y(x) = -3/(2+4x) + 1 2.) Solving inequalities is an algebra skill, the hard part is understanding that a gradient is just the derivative. So, you have the inequality... 1/(1+2x)^2 < 1/3 Find asymptotes... (1+2x)^2 = 1/3 1 + 2x = sqrt(1/3) x = [sqrt(1/3) - 1] / 2 = -0.21132486 try x = -1 and x = 0... dy/dx(-1) = 1/(1-2)^2 = 1/(-1)^2 = 1 > 1/3 (so x less than this point won't work) dy/dx(0) = 1/(1+0)^2 = 1 > 1/3 (so x greater than this point won't work) So, in other words, dy/dx is always greater than 1/3, so there are no x values where the gradient of the curve is less than 1/3.