Show that the equation $ x^4 + 4x + c = 0 $ has at most two real roots.
all right, Show that equation except forthe path for X Plus C is equal to zero has at most two real route. Okay, so the first thing we can do is we're going to take the derivative off f this function right here. So the derivative is simply for X cube plus four engine Souza Constant that zero and Ciro di Dio Andi Now we're going to pull out the factory for that is very bad for you for X cube plus one and we can see at this function here equals zero when X is equal to negative one. So since this is the derivative and we know that X is equal to negative one, we know that at X equals negative one some sort of some sort of extreme err occur So some sort of local men or local max occured because if you look at it, some sort of increasing or decreasing hast occur around X equals negative one. That's what it means. The derivative that Qianjin Slope, the slope at it is equal to zero. So some sort of movement has to occur for this function has to occur in order for this to be in a minute, Max, and that's what this tells us. And so there's a serum that states that if after backs if after Max has and number of local extreme int local extreme then at most disfunction then after make that most has n plus one zero has that most end plus one zero since we found one end that has a pop has a local extreme A. We know for a fact that by this dear um, that and that there's only end plus one possible zero. And since we have one isn't musical toe one. In our case, we know that there is n plus one possible Broots, which is too. This proves that there's two part at, at most two real groups in the function.