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Numerade Educator



Problem 20 Medium Difficulty

Show that the equation has exactly one real root.
$ x^3 + e^x = 0 $




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Video Transcript

Okay, Now we are being asked to show that the equation has exactly one root. The equation is x x cubed, plus the F X equals zero. And we're trying to show we're trying to prove that this function, that there's only one value that gives us ex Cuba's here practical zero. And so there are many different ways to prove this. And one way to prove I'm going to use involved limit. So if we take the limit as X goes to negative infinity mhm of X Cube plus FX, what happened? So what actually happens is we get a really, really big negative infinity number because at execute growth, um, it was just for a negative 100 cube. That is still negative. So give us an even bigger negative infinity numbers. So we're gonna label that is negative infinity, and then we're gonna get plus e to the negative infinity. Now, recall that this is an exponent should eat any number raised to a negative exponents means you put it under you put it under his nominator. So this so e to the negative infinity become one over e to the infinity and one over a very large number approaches zero. So this approach is negative. Infinity now the limit as X goes to positive infinity of X cube plus e to the X. It just simply infinity plus infinity and infinity Plus infinity is still infinity. It's still a very large number. So now now you can see something very interesting happening. So when the number is very, very, very, very negative, it's negative Infinity. And this one is very, very, very positive. It's positive infinity. So that means that there is a point such that, um between negative infinity and infinity that this function has to cross the x axis. So there has to be a point where this is equal to zero. And this is true because we know this by the intermediate value zero. And we know this because this function executes but the effect is a continuous function. X cubed is a cubic funding is continuous and expert and each of the excess and expanding function which is also continuously and on all real numbers. So by the enemy of value theorem, we know for a fact that there is one group, but not that we know that there is at least a route, not necessarily one route to prove that there is only one route we have to take the derivative of X Cube plus 80 X, And the reason why is this tells us a lot about how the function is behaving and so three X square plus e. T. X. Because that's the derivative. And this derivative is always greater than zero is always increasing because X square. Even if you put a negative number, it is always positive. And so since this function is always positive, we know that it's always increasing. It never decreases. It always rises, it always goes up. And so when this is that negative infinity, we know that this has to cross the X axis ones because it's always going to be increasing. And so that proves that there's only one route. So by taking derivative, we prove that there's only one group because this function is always increasing on all domain, and so that proves it so. In general, when you're trying to prove whether the equation has one root, you were generally probably going to use the enemy value at some point to prove that there is even a route so That's a good That's a good start import and then the second. And then when you want to prove that there is only one route, there is a variety of way. But you can take a very easy way to check is to take the derivative and see how that function is increasingly decreasing on its in Vall. And I can tell you a lot about it, um, about his behavior, and I can tell you about whether there's one route group or give you some sort of sense of idea of how many routes they are. So in this case, we know that there was only one room because the function is always increasing and we know that there is a group due to the enemy factor I remember.