Find the exact points of intersection of the two graphs of f(x) = x^2 + 2x + 2 g(x) = x + 2
This question is pretty simple when you understand the basic rule that two graphs intersect when the equations are equal to each other. So, to begin, you'd set the equation x^2 + 2x + 2 equal to x + 2. This would give you x^2 + 2x + 2 = x + 2. The next thing you want to do is make your whole equation equal to zero so that it's easier to solve. You'd subtract 2 from both sides to get the equation x^2 + 2x = x. Then you can subtract x from both sides to get the equation x^2 + x = 0. Now that everything is equal to zero, you can factor the equation. In this case, you can factor out an x to get x(x + 1) = 0. Now, because we're trying to figure out the values of x that make this equation equal to zero and because any number multiplied by zero is equal to zero, we need to find when each of these parts of the equation are equal to zero. The first part, x, is easy because x is equal to zero when x = 0. The second part takes a bit more thought. For that, you need to find when x + 1 is equal to zero. Here, it happens when x = -1. So, you've just found two x values where the two graphs intersect: at x = 0 and x = -1
This question comes from a BC Calc practice AP test: Determine dy/dt given y = x^2 + 4x x = cos(3t)
The first thing to do is to take the derivative of y with respect to x. However, because x is a function in this case, and not just a variable, we cannot assume that the derivative of x is equal to one. So we need to account for the derivative of x when we're taking the derivative of y. This will give you dy/dx = (2x)(x') + 4(x'), where x' is the derivative of x. Now that you have the derivative of y with respect to x, you can plug the function x into dy/dx for all x. This will give you the equation (2cos(3t))(x') + 4(x'). Then, you take the derivative of x with respect to t. That gives you the equation dx/dt = -3sin(3t). Now you can plug in that equation for x' in the other equation. So, you have (2cos(3t))(-3sin(3t)) + 4(-3sin(3t)), which is dy/dt
This question comes from an AB Calc practice AP test: Given f(x) = 2x + 2 and g(x) = -2x + x^2, find f(g(1)).
The first thing you have to do given this problem is figure out what g(1) is, which you would do by plugging 1 in for x in the equation g(x) = -2x + x^2. This would give you the equation g(1) = -2(1) + (1)^2. After simplifying, you get g(1) = -2 + 1, which = -1. Now, because the problem asked for f(g(1)), we're not done yet. You've just found that g(1) = -1, so you can substitute -1 in for g(1) to get f(-1). In order to solve f(-1), you do the same thing you did earlier with the g(x) function. You plug in -1 for x into the equation f(x) = 2x + 2. This gives you the equation f(-1) = 2(-1) + 2, which you can simplify to be f(-1) = -2 + 2, which = 0. So you've found that f(g(1)) = 0.