a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$ f(x)=x^{1 / 3}(x+8) $$

Answer

\documentclass{article} \usepackage{amsmath , amssymb ,amsthm} \usepackage[utf8]{inputenc} \begin{document} a) To find out intervals on which the function is increasing or decreasing we need to find out the sign of the derivative function. Because the derivative of a function shows its slop in each point. When the slop is positive the function is increasing and when the slop is negative the function is decreasing.\\ We have $$f(x) = x^{4/3}+8x^{1/3} $$ And rthe derivative of this function is $$f'(x) = \frac{4}{3}x^{1/3}+\frac{8]{3}x^{-2/3} $$ The derivative function is always positive and to make sure about this result we find $f'(x)=0$ points, which are -2 and 0. -2 is not acceptable because it cauese a root of negative number. The only root is 0 and for $x>0$, $f'(x)$ is positive. \\ The domain of the original function is $[0,\inf)$ so the function is positive in all its domain. \\ b) As the only root of $f'(x)$ is 0 and $f'(0)=\inf$ then 0 is saddle point and the function does not have any minimum or maximum. \end{document}

So we have the function f of X equals X to the third power to the 1/3 power times expressing and to find where it's increasing and decreasing, we need to take the first derivative so f prime of X equals. Now if we use the product rule for, um differentiation. First, we take the derivative of X to the 1/3 just 1/3 times x to the negative 2/3 times X plus eight plus X to the 1/3 times the derivative of X plus eight, which is just one. We can simplify that a little. So we get, um, X plus eight over three X to the 2/3 um, plus X to the 1/3. So So now what we can do is find the critical values of F of f of X and the critical values of function. Um, are where, um, the first derivative either equals zero or undefined. So if we're looking for when f prime of X is undefined, that's gonna be when the denominator three x to the 2/3 equals zero. And clearly this occurs when X is equal to zero. So we know that's one critical value at X equals zero. Now we have to find when the first derivative is equal Teoh zero. So if we multiply out by three x to the 2/3 we get X Plus eight Pleiss, um three X equals zero records three x to the 2/3 times X to the 1/3 2 3rd less 1/3. It's just, um X to the first, um so we simplify that we get four X plus eight equals zero. So clearly X equals negative, too. Tracked by eight on both sides. And divide by Force X equals negative, too. So we have a critical value at X equals zero, and X equals negative too. So now what we can do is make a number line in order to show how f prime of X behaves around these critical values. So if we draw a number line, um, with our critical values on it, we have zero. And negative too. Now we want to find, um, we want to find where weather F prime of X is negative or positive, um, around thes critical values. So if we use the test value in each of these intervals, we can, um, we confined um, whether it's negative or positive. So if we use negative nine for this value, a negative nine turns out to be a friendly number when we're working with some one to the U third roots and whatnot. So if we look for F prime of, um, negative nine, that becomes negative nine plus eight over three times the third roots of negative nine. Swear to the third roots of 81. Um, because, uh, X to the 2/3 is just same as 1/3 root of X squared. Plus negative nine to the 1/3 power equals negative nine plus eight. That's negative. One over. Que brute of um, 81 is, um, if we use a calculator, the cube root of 81 is just going to be, um, about 4.32 It's negative. One over 4.3 will say. Plus, um, negative nine. The third root of negative nine, which is negative 2.8 So clearly a negative number, plus another negative number. This is gonna be less than zero. Um, so we have that when X is less than negative, too. F prime of X is negative. Um, Now, between negative two and negative one we can. So if you use the test value, X equals negative one f prime of negative one is equal to negative one plus eight that seven times. Um three time. Oh, I just noticed. Um, this should be 4.3 times three. Um, it doesn't actually affect whether or not it is negative or positive, but, um, just for clarity's sake. So back to F prime of negative one. We have 7/3 times, um, the cube root of one or negative one squared, which is just one plus the cube root of negative one, which is There we go and that's equal to seven. Que brood of one is just ones that 7/3. Plus, um, que brood of negative one is negative one. So we have 7/3, minus one or 7/3 minus 3/3, which gives us for over three, which is positive. So we know from 2 to 0 or from negative 2 to 0. That's positive. Then, if we want to use a test value for X is greater than zero, we can just use one. So we have f prime of one is one plus eight is nine, three times the cube root of one squared again. That's just gonna give us one. Um, plus, the cube root of, um, one is again just gonna give us one. So we have 9/3 plus one, which, um, clearly is greater than zero. So we have that. It's also positive when X is greater than zero. So now we know the intervals of when F prime of X is negative and positive. So does that actually tell us about ffx? Well, when the first derivative of a function is negative, it means that the function is decreasing on that interval. Um so f of X is decreasing on that interval, then it gets too negative, too, and it starts increasing again. And when it gets to zero, it's still increasing. So we know the interval in which the function is decreasing is negative. Infinity too, um, to Heurtaux negative too. And it's increased saying on, um negative 2 to 0 and zero to positive infinity. Now, if we look at when it's decreasing and increasing, we see that right here at X equals negative two. When it goes from decreasing the increasing we have a local minimum. So because it's at the very bottom of this little valley here it goes from all the way. Decreasing toe back to increasing with local minimum at X equals two X equals negative, too.

## Discussion

## Video Transcript

So we have the function f of X equals X to the third power to the 1/3 power times expressing and to find where it's increasing and decreasing, we need to take the first derivative so f prime of X equals. Now if we use the product rule for, um differentiation. First, we take the derivative of X to the 1/3 just 1/3 times x to the negative 2/3 times X plus eight plus X to the 1/3 times the derivative of X plus eight, which is just one. We can simplify that a little. So we get, um, X plus eight over three X to the 2/3 um, plus X to the 1/3. So So now what we can do is find the critical values of F of f of X and the critical values of function. Um, are where, um, the first derivative either equals zero or undefined. So if we're looking for when f prime of X is undefined, that's gonna be when the denominator three x to the 2/3 equals zero. And clearly this occurs when X is equal to zero. So we know that's one critical value at X equals zero. Now we have to find when the first derivative is equal Teoh zero. So if we multiply out by three x to the 2/3 we get X Plus eight Pleiss, um three X equals zero records three x to the 2/3 times X to the 1/3 2 3rd less 1/3. It's just, um X to the first, um so we simplify that we get four X plus eight equals zero. So clearly X equals negative, too. Tracked by eight on both sides. And divide by Force X equals negative, too. So we have a critical value at X equals zero, and X equals negative too. So now what we can do is make a number line in order to show how f prime of X behaves around these critical values. So if we draw a number line, um, with our critical values on it, we have zero. And negative too. Now we want to find, um, we want to find where weather F prime of X is negative or positive, um, around thes critical values. So if we use the test value in each of these intervals, we can, um, we confined um, whether it's negative or positive. So if we use negative nine for this value, a negative nine turns out to be a friendly number when we're working with some one to the U third roots and whatnot. So if we look for F prime of, um, negative nine, that becomes negative nine plus eight over three times the third roots of negative nine. Swear to the third roots of 81. Um, because, uh, X to the 2/3 is just same as 1/3 root of X squared. Plus negative nine to the 1/3 power equals negative nine plus eight. That's negative. One over. Que brute of um, 81 is, um, if we use a calculator, the cube root of 81 is just going to be, um, about 4.32 It's negative. One over 4.3 will say. Plus, um, negative nine. The third root of negative nine, which is negative 2.8 So clearly a negative number, plus another negative number. This is gonna be less than zero. Um, so we have that when X is less than negative, too. F prime of X is negative. Um, Now, between negative two and negative one we can. So if you use the test value, X equals negative one f prime of negative one is equal to negative one plus eight that seven times. Um three time. Oh, I just noticed. Um, this should be 4.3 times three. Um, it doesn't actually affect whether or not it is negative or positive, but, um, just for clarity's sake. So back to F prime of negative one. We have 7/3 times, um, the cube root of one or negative one squared, which is just one plus the cube root of negative one, which is There we go and that's equal to seven. Que brood of one is just ones that 7/3. Plus, um, que brood of negative one is negative one. So we have 7/3, minus one or 7/3 minus 3/3, which gives us for over three, which is positive. So we know from 2 to 0 or from negative 2 to 0. That's positive. Then, if we want to use a test value for X is greater than zero, we can just use one. So we have f prime of one is one plus eight is nine, three times the cube root of one squared again. That's just gonna give us one. Um, plus, the cube root of, um, one is again just gonna give us one. So we have 9/3 plus one, which, um, clearly is greater than zero. So we have that. It's also positive when X is greater than zero. So now we know the intervals of when F prime of X is negative and positive. So does that actually tell us about ffx? Well, when the first derivative of a function is negative, it means that the function is decreasing on that interval. Um so f of X is decreasing on that interval, then it gets too negative, too, and it starts increasing again. And when it gets to zero, it's still increasing. So we know the interval in which the function is decreasing is negative. Infinity too, um, to Heurtaux negative too. And it's increased saying on, um negative 2 to 0 and zero to positive infinity. Now, if we look at when it's decreasing and increasing, we see that right here at X equals negative two. When it goes from decreasing the increasing we have a local minimum. So because it's at the very bottom of this little valley here it goes from all the way. Decreasing toe back to increasing with local minimum at X equals two X equals negative, too.

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