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. $$ f(x)=x-6 \sqrt{x-1} $$

All right, so four problem 31 function given is X minus six times square root of X minus one. So for the first part, I mean to take the derivative of dysfunction, which is one finest six times one over two times square root X minus one. And because of their change, will have to time. We have two times this direction by the derivative off the inside function, which is just one. So we can, um, just a ride this fraction here now. So this is a derivative. Now we need to let this drew to boot to be positive. And so, Dan Cues one minus, re divided by X minus one, which is bigger than zero and correspondingly, and we can have three divided by you were rude. X minus one is smaller than one. Now, here. We need to, uh we need to make a note. Note is, since x minus one is under the square root, so x minus one has to be positive. So that means, uh, domain of dysfunction is from one to infinity. Okay, So that means X cannot be won. Hence we can multiply square root, X minus one on both sides off busying quality. Therefore, this gives square root off X minus one bigger than three. Now we take the square on both eyes. Now we have X minus one bigger than nine, which means access bigger than 10. Therefore, our function is increasing the interval from 10 to infinity and it is decreasing from 1 to 10. So we're done Helpful part B. I need to find the local extreme. So we observed the the intervals for increasing and decreasing. We can simply conclude that app 10 will be and loco extreme value. So the next thing is to check whether this function value is is the is the absolute absolute, absolute extreme value. Now we try to draw the in order to solve the problem. We try to draw the graph of dysfunction by using these these information Thean DeVos for increasing and decreasing. So what we have is from 1 to 10 to 70 and from 1 to 10 function will be decreasing and from 10 to infinity function will be increasing. Okay, so that means r f one. The end point might be, um wait. Um yet R f one could be ah, could be. Could be an absolute extreme. We check the function value that one will be, um, will be just the one. If we check the banks value off at 10 this will be 10 minus 18 which is an active aid. So that is our the absolute absolute minimum of dysfunction. And remember, our function is from the are a function domains from one to infinity. So that's the only thing that we need to check We need to track in this case. Okay, so I'll just, uh, circle this function value. This is absolute extreme, and we're done.

## Discussion

## Video Transcript

All right, so four problem 31 function given is X minus six times square root of X minus one. So for the first part, I mean to take the derivative of dysfunction, which is one finest six times one over two times square root X minus one. And because of their change, will have to time. We have two times this direction by the derivative off the inside function, which is just one. So we can, um, just a ride this fraction here now. So this is a derivative. Now we need to let this drew to boot to be positive. And so, Dan Cues one minus, re divided by X minus one, which is bigger than zero and correspondingly, and we can have three divided by you were rude. X minus one is smaller than one. Now, here. We need to, uh we need to make a note. Note is, since x minus one is under the square root, so x minus one has to be positive. So that means, uh, domain of dysfunction is from one to infinity. Okay, So that means X cannot be won. Hence we can multiply square root, X minus one on both sides off busying quality. Therefore, this gives square root off X minus one bigger than three. Now we take the square on both eyes. Now we have X minus one bigger than nine, which means access bigger than 10. Therefore, our function is increasing the interval from 10 to infinity and it is decreasing from 1 to 10. So we're done Helpful part B. I need to find the local extreme. So we observed the the intervals for increasing and decreasing. We can simply conclude that app 10 will be and loco extreme value. So the next thing is to check whether this function value is is the is the absolute absolute, absolute extreme value. Now we try to draw the in order to solve the problem. We try to draw the graph of dysfunction by using these these information Thean DeVos for increasing and decreasing. So what we have is from 1 to 10 to 70 and from 1 to 10 function will be decreasing and from 10 to infinity function will be increasing. Okay, so that means r f one. The end point might be, um wait. Um yet R f one could be ah, could be. Could be an absolute extreme. We check the function value that one will be, um, will be just the one. If we check the banks value off at 10 this will be 10 minus 18 which is an active aid. So that is our the absolute absolute minimum of dysfunction. And remember, our function is from the are a function domains from one to infinity. So that's the only thing that we need to check We need to track in this case. Okay, so I'll just, uh, circle this function value. This is absolute extreme, and we're done.

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