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Determine the intervals on which the given function $f$ is increasing and the intervals on which $f$ is decreasing. $$ f(x)=x^{2 / 3}-2 x^{1 / 3} $$

So without a calculator, what you want to do is find the derivative of this function defined as X squared minus one cubed. And with the derivative you just have to make sure you do the chain rule where you move the three this three out front. Um What you want to make sure you do is leave the inter functional alone to the second power. Now from the power rule Times the derivative of the inside though, which would be two x. Um And what you need to do is examine where the derivative is equal to zero. So in this problem you have to pay attention to a few things because this piece is squared. Um I'll show the arithmetic here that you would set X squared minus one equal to zero. So you add one over and acts could be plus or minus one but they're each actually a multiplicity of to the same thing with this zero X equals zero. So if you were to make a sign chart with this which is pretty common. And you're trying to see on the number line negative on zero and 1 where are you going to get you know the derivative is negative. I just think about how to the left of zeros are negative values and then to the right of zeros are going to be positives. And I'm looking at where the dirigible equal will be a positive. So a negative times a negative times a negative times a negative times a negative goopy and negative. And then same thing here we have three negatives will be a negative. Everything to the right here will be a negative times the name is positive times the other positives. So if you go back to the original problem um I would just say because it is a polynomial that from since the derivative is negative. Um that means the graph is decreasing. I would just say that that is going from negative infinity until zero and then from there the derivative is positive. I misspelled decreasing so then it's increasing From 0 to Infinity. And I know some teachers might put a little break and negative one or positive one. Um So you might have a teacher that does that. I don't because I think if it's decreasing to the left and the right then we can just say it's decreasing throughout. So like the graph and look like that. And same thing with the positive, the positive goes up but it doesn't really stop increasing. But that's my opinion. Yeah.

I want to just quickly point out with this one. Um the function to find As 1 -1 to the one third power. Um that the domain on this is all reels because just a reminder that is the same thing as one minus the cube root of X. And you just may remember what the Q bert function looks like. Uh except this one with the way it shifted is shifted up one and it's negated. So it's actually going this way. So you probably already know the answer. But what we want to show uh in calculus is that the derivative of this will be negative one third X to the negative two thirds power. Uh So in other words, that's the same thing as negative of one over three cube root of X squared. And when you consider this ever equaling zero, It's impossible to equal zero. So there are no critical numbers. Uh If you're confused, why? Uh X squared? Uh no matter what you put in for X once you square becomes a positive um and then one will never equal zero. Negative one will never equal zero. So no matter what you plug in, you're going to have a negative Divided by a positive negative one divided by positive, which is always negative. And if that's the case, you know, if the derivative is negative, that implies that we are always decreasing on the domain which is from negative infinity to infinity, it's never increasing

A student that has access to a graphing calculator and I don't even think you need a graphing calculator on this one too. Be able to identify what this graph will look like. Um Because in freak out you learn about how to factor out X squared um minus three X squared. There we go. So if you wanted to, if you think back to pre calculate might even rewrite the problem like this and you can say okay X equals zero, I have a positive leading coefficient. So it's got to have this end behavior and then exit goes 123 is another zero and it crosses through. Um So that might help you identify if you find the derivative of the function F using your power rule. Yeah. And then factor out a three x. So you'd be left with X -2. You can see that the derivative changes At zero and x equals two from a positive to a negative value. And knowing this parent function um would help you I think tell me that it's increasing where the derivative is positive And that's on the interval from negative infinity to zero. Um I like to use parentheses. I know some people do brackets. Uh huh. Um And you can verify that this is correct because if you find let's say you plug in a negative number in here. Well negative once you square it's going to be a positive. Um And then this will be a positive. So positive plus another positive will give you a positive derivative. You can test it with numbers bigger than two as well and therefore it's decreasing everywhere else. Or the derivative is negative, And that happened on the interval from 0 to 2.

Uh If you look at this function f of X equals X cubed. A good student in pre calculus even answer it to probably knows this is what the graph looks like and it's always going up into the right. Um So you might say that it's increasing on the domain or on all reels from negative infinity to infinity. Now you might also uh have a teacher though is say, well technically it actually goes zero. The slope is zero. Um I like this answer the most, but I will point out that you might have a teachers say that they don't want you to include zero. So let me discuss what's happening there is it's increasing because you write this in blue, the derivative is positive. All right everywhere. Um So if you did the derivative of this function, You will get three x squared. And because X, no matter if X is negative or positive when you square it, you're going to get a positive number. Uh the only argument would be at X equals zero. Uh So that's your reason why you are increasing. Um and the reason why I would include zero, even though yes, the derivative equal zero. There is, it's increasing to the left and it's increasing to the right. So I would say it's increasing throughout. So there's a little discussion for you as you move forward