How do i find the Crit.#'s and PIPS here | Mathematics | Page 1

lofty cloud Mar 18, 2024, 3:05 AM

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gusty cipherBOT Mar 18, 2024, 3:05 AM

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lofty cloud

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versed terrace Mar 18, 2024, 3:07 AM

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lofty cloud

Well, I think a good place to start would be reading the instructions.

lofty cloud Mar 18, 2024, 3:08 AM

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i did i have to find the crit.#s and pips in order to do the second derivative test

versed terrace Mar 18, 2024, 3:08 AM

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lofty cloud i did i have to find the crit.#s and pips in order to do the second derivative t...

"PIPs"?

lofty cloud Mar 18, 2024, 3:09 AM

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Potential Inflection Points

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thats what our instructor called it

versed terrace Mar 18, 2024, 3:09 AM

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...okay, so how do you do that?

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Like, you were taught that, right?

lofty cloud Mar 18, 2024, 3:11 AM

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yeah we equaled the equation to 0 and than factored. than lets say we had (x - 2)(x + 3) then our crit #s or our PIPs would be x = 2 and -3

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but i dont know how i would factor out here or how i should find them

versed terrace Mar 18, 2024, 3:11 AM

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Okay, the factoring part is not important.

lofty cloud Mar 18, 2024, 3:11 AM

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like is there another method

versed terrace Mar 18, 2024, 3:12 AM

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What's important is that you set the derivative equal to 0, and then solve for x.

lofty cloud Mar 18, 2024, 3:14 AM

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so i would just use square root of 1 during the test?

versed terrace Mar 18, 2024, 3:14 AM

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lofty cloud so i would just use square root of 1 during the test?

...what's the square root of 1?

lofty cloud Mar 18, 2024, 3:15 AM

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1

versed terrace Mar 18, 2024, 3:15 AM

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Or?

lofty cloud Mar 18, 2024, 3:15 AM

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-1

versed terrace Mar 18, 2024, 3:15 AM

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What's the domain of h?

lofty cloud Mar 18, 2024, 3:15 AM

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what do you mean domain of h

versed terrace Mar 18, 2024, 3:16 AM

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...I mean the values of x over which h is defined.

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In real numbers.

lofty cloud Mar 18, 2024, 3:17 AM

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mb still not sure what you mean

versed terrace Mar 18, 2024, 3:17 AM

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Try to calculate h(-1) for me, please.

lofty cloud Mar 18, 2024, 3:18 AM

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oh they both equal to 0

versed terrace Mar 18, 2024, 3:18 AM

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No.

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h(-1), not h'(-1).

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Well?

lofty cloud Mar 18, 2024, 3:20 AM

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ln-1 is just coming up as an error

versed terrace Mar 18, 2024, 3:20 AM

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Let's think about why that might be. What's ln?

lofty cloud Mar 18, 2024, 3:21 AM

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it sa natual log or in other works log base e

versed terrace Mar 18, 2024, 3:21 AM

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Right. So if ln(-1) = k, then -1 = e^k, right?

lofty cloud Mar 18, 2024, 3:22 AM

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yeah

versed terrace Mar 18, 2024, 3:22 AM

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Can e^k ever be negative if k is a real number?

lofty cloud Mar 18, 2024, 3:23 AM

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no right?

versed terrace Mar 18, 2024, 3:24 AM

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Right. e^k is always positive for real k. Therefore the range of e^x is (0, inf), and since ln(x) is its inverse, it has that same interval as its domain.

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Trying to take a logarithm of a non-positive number doesn't give us a real answer.

lofty cloud Mar 18, 2024, 3:25 AM

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oh so because of that for what i was doing earlier -1 isnt in the domain of h

versed terrace Mar 18, 2024, 3:25 AM

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Right.

lofty cloud Mar 18, 2024, 3:25 AM

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becuase -1 cant exist with ln

versed terrace Mar 18, 2024, 3:25 AM

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And let's take a moment to just clarify why we're so interested in solutions to h'(x) = 0.

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We're looking for local extrema, right?

lofty cloud Mar 18, 2024, 3:26 AM

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yeah

versed terrace Mar 18, 2024, 3:26 AM

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What are those?

lofty cloud Mar 18, 2024, 3:26 AM

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rel max or min

versed terrace Mar 18, 2024, 3:26 AM

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Okay, but what are those?

lofty cloud Mar 18, 2024, 3:27 AM

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when the slope is 0

versed terrace Mar 18, 2024, 3:27 AM

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Why?

lofty cloud Mar 18, 2024, 3:27 AM

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the line goes from neg to pos or pos to neg, its a curve

versed terrace Mar 18, 2024, 3:28 AM

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Right, but what I'm trying to guide you towards is the visual intuition.

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A local maximum is, literally, a value of the function that is higher than any other point around it.

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A local minimum is a value of the function that is lower than any other point around it.

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So in order to be a maximum, the curve has to climb as it approaches the maximum, level off at the maximum, and then dive as it leaves the maximum.

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If it's continuous at the maximum, at least.

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So, like - wait.

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If you kinda already knew all this, why were you so confused about how to find critical points?

lofty cloud Mar 18, 2024, 3:33 AM

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i learned this on wednesday and friday and have homework due tommorow, i kinda put it off till right now. And im also kind of confused on what i learned for extrema, first and second deriv test, and curve sketching

versed terrace Mar 18, 2024, 3:34 AM

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Okay, well, like we just talked about, (potential) extrema are the points at which the derivative is zero because, with the exception of discontinuity, if the derivative wasn't 0 at that point, it would be impossible for it to be an extremum.

lofty cloud Mar 18, 2024, 3:36 AM

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yeah that makes sense

versed terrace Mar 18, 2024, 3:37 AM

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So we found x = 1, so what do we do now?

lofty cloud Mar 18, 2024, 3:37 AM

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do the test and find extrema and inf. pts.

versed terrace Mar 18, 2024, 3:37 AM

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...do what test, how, on what?

lofty cloud Mar 18, 2024, 3:39 AM

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oh do the test with x = 1, but from what i see here isnt this just an inflection point since it stays positive, and i can send a picture of what the test looks like

versed terrace Mar 18, 2024, 3:39 AM

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...what "stays positive"?

lofty cloud Mar 18, 2024, 3:41 AM

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versed terrace Mar 18, 2024, 3:41 AM

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lofty cloud

That doesn't help me.

lofty cloud Mar 18, 2024, 3:42 AM

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what stays positive is the slope

versed terrace Mar 18, 2024, 3:43 AM

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Okay, I'm like 90% sure you've confused yourself, but the only way that I can show you that is if you're more explicit about what you did.

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Like, all I see is a chart. You didn't tell me how you drew the chart or filled it in.

lofty cloud Mar 18, 2024, 3:48 AM

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so after i found the derivative, i set the equation to zero and got the crit #. when i made the chart the numbers with the line going up are the crit #s and the numberst to the right and left are just the direct numbers next to them(if there are multiple numbers you put 0 in the middle) after the i plugged in the equation to the chart and checked where it was positive and where it was negative

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oh i just saw that i did 1 instead of x - 1/x

versed terrace Mar 18, 2024, 3:48 AM

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"Plugged in the equation to the chart" is not... a thing.

versed terrace Mar 18, 2024, 3:49 AM

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lofty cloud oh i just saw that i did 1 instead of x - 1/x

Yeah, it would've been something like that, but also this is the first derivative test, not the second.

lofty cloud Mar 18, 2024, 3:51 AM

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oh

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i finished setting the second derivative to 0 and got √-1

versed terrace Mar 18, 2024, 3:52 AM

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That's also not the second derivative test.

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The second derivative test is when we evaluate the second derivative at the zeroes of the first derivative.

versed terrace Mar 18, 2024, 3:53 AM

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lofty cloud i finished setting the second derivative to 0 and got √-1

But this does mean that there's no points of inflection.

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Well, again, no real points of inflection.

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Since sqrt(-1) = i.

lofty cloud Mar 18, 2024, 3:54 AM

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honestly sorry but im so confused right now

versed terrace Mar 18, 2024, 3:55 AM

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Okay, well let's relax and take it slow.

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What does the second derivative measure?

lofty cloud Mar 18, 2024, 3:56 AM

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extrema, concavity, and inflection points

versed terrace Mar 18, 2024, 3:57 AM

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Okay, it doesn't measure extrema.

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Concavity was the answer I was looking for.

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That includes inflection points, since an inflection point is just the point at which concavity changes.

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First derivative measures slope, second derivative measures concavity.

lofty cloud Mar 18, 2024, 3:59 AM

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oh okay so we find extrema with the first by seeing when the slope changes, and concavity/inflection with the second

versed terrace Mar 18, 2024, 3:59 AM

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Right.

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And the second derivative test tells us which kind of extremum we're looking at.

lofty cloud Mar 18, 2024, 4:00 AM

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yeah if it faces down or up right?

versed terrace Mar 18, 2024, 4:00 AM

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Because what does it mean when the second derivative is positive?

lofty cloud Mar 18, 2024, 4:01 AM

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its facing up/ concave up

versed terrace Mar 18, 2024, 4:01 AM

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So if h'(a) = 0 and h''(a) > 0?

lofty cloud Mar 18, 2024, 4:01 AM

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then its concave up

versed terrace Mar 18, 2024, 4:02 AM

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And?

lofty cloud Mar 18, 2024, 4:02 AM

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its a min

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at a

versed terrace Mar 18, 2024, 4:02 AM

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Right. And if h'(a) = 0 and h''(a) < 0?

lofty cloud Mar 18, 2024, 4:02 AM

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then its a max at a and concave down

versed terrace Mar 18, 2024, 4:02 AM

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Right. More specifically it's a max at a because it's concave down.

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So what if h'(a) = 0 and h''(a) = 0?

lofty cloud Mar 18, 2024, 4:03 AM

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isnt that just a line like h(x)=a

versed terrace Mar 18, 2024, 4:04 AM

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Not necessarily.

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Consider f(x) = x^3.

lofty cloud Mar 18, 2024, 4:06 AM

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oh so than its an inflection point where it goes from pos to 0 to pos or neg to 0 to neg

versed terrace Mar 18, 2024, 4:06 AM

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Right. Because what "concavity" really is is the rate of change of the slope.

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"Concave down" is when the slope is decreasing, "concave up" is when it's increasing.

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So if the slope at a point is 0, and the slope is increasing at that point, that means that to the right of that point the slope is greater than 0.

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And to the left of that point the slope is less than zero, because it increased to zero.

lofty cloud Mar 18, 2024, 4:08 AM

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yeah

versed terrace Mar 18, 2024, 4:08 AM

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And if the slope at a point is 0 and it's decreasing, that means it was more than 0 on the left and less than 0 on the right.

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But then when we look at f(x) = x^3, we see a case where the slope is 0 and the change in slope is also 0, and in this case it turns out to be because that's the point that the change in the slope goes from decreasing to increasing; an inflection point.

lofty cloud Mar 18, 2024, 4:10 AM

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okay that makes sense

versed terrace Mar 18, 2024, 4:11 AM

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But a 0 in the second derivative isn't always an inflection point, because, y'know, what's the second derivative of f(x) = a?

lofty cloud Mar 18, 2024, 4:11 AM

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0

versed terrace Mar 18, 2024, 4:11 AM

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Right, but it's just a horizontal line. It has no slope, much less change in slope.

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So anyway, going all the way back to the original question.

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We have h'(x) = 0 when x = 1.

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So we know x = 1 is something. How do we figure out what it is?

lofty cloud Mar 18, 2024, 4:13 AM

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we do the test to see what we get at that point

versed terrace Mar 18, 2024, 4:14 AM

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lofty cloud we do the test to see what we get at that point

...okay, can you use words other than "do the test"?

lofty cloud Mar 18, 2024, 4:15 AM

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use the 1st derivative test to find which extrema it is

versed terrace Mar 18, 2024, 4:15 AM

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Is that what the question says to do?

lofty cloud Mar 18, 2024, 4:16 AM

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it says exterma, but than also inflection points which can be gotten from the 2nd derivative test

versed terrace Mar 18, 2024, 4:16 AM

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It says use the second derivative test.

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That's what it says.

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Use the second derivative test to find the extrema.

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That is what the question says to do.

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And I just went to great pains to explain in excruciating detail exactly what that means and why it works.

lofty cloud Mar 18, 2024, 4:18 AM

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nono im pretty sure i understand this now

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i really am thankful and your explenation has helped

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i got to use the 2nd derivative test to find the extrema and the inflection points

versed terrace Mar 18, 2024, 4:19 AM

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By doing what?

lofty cloud Mar 18, 2024, 4:20 AM

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by finding if h''(a) < 0 or h''(a) > 0

versed terrace Mar 18, 2024, 4:20 AM

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And what's a in this context?

lofty cloud Mar 18, 2024, 4:21 AM

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1

versed terrace Mar 18, 2024, 4:22 AM

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So what result do we get?

lofty cloud Mar 18, 2024, 4:23 AM

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i got 2

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so h''(1) = 2 > 0

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so its concave up

versed terrace Mar 18, 2024, 4:24 AM

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Which means?

lofty cloud Mar 18, 2024, 4:26 AM

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rel. min. at h(1), dec. from (-inf, 1), and inc. from (1, inf)

versed terrace Mar 18, 2024, 4:26 AM

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...no.

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First of all, we talked about the function's domain, right?

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Second of all, the second derivative has no real zeroes, meaning that concavity doesn't change direction.

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Thirdly, why are you even talking about the intervals of decrease or increase? Those weren't part of the question.

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But also, concavity never changes direction. That means that the slope of the function never decreases, so that's not just a relative minimum, it's the absolute minimum.

lofty cloud Mar 18, 2024, 4:33 AM

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oh okay, and what do you mean when you said that concavity nevery changes direction and that slope never decreases. Isnt that what happenes on the left of concave up and on the right of concave down?

versed terrace Mar 18, 2024, 4:35 AM

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lofty cloud oh okay, and what do you mean when you said that concavity nevery changes direct...

Okay, look, the function is concave down at x = 1, right?

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That means the slope is increasing, right?

lofty cloud Mar 18, 2024, 4:35 AM

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ok

versed terrace Mar 18, 2024, 4:35 AM

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But the slope is always increasing, is my point.

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It starts negative, increases to 0, then increases past 0.

lofty cloud Mar 18, 2024, 4:36 AM

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oh ok

versed terrace Mar 18, 2024, 4:36 AM

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Like, when is h''(x) < 0?

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Only when x^2 < -1, which is impossible for real x.

#How do i find the Crit.#'s and PIPS here