find-the-points-on-the-curve-y2-x33-x2-12-x1-where-the-tangent-is-horizontal

Name: find the points on the curve y2 x33 x2 12 x1 where the tangent is horizontal
Uploaded: 2020-06-05
Duration: 6 min 2 s
Channel: Numerade
Description: Find the points on the curve $y=2 x^{3}+3 x^{2}-12 x+1$ where the tangent is horizontal.

Question

Find the points on the curve $y=2 x^{3}+3 x^{2}-12 x+1$ where the tangent is horizontal.

Kira Schwander · Accepted Answer

Hello. We&#x27;re going to be working with the curve. Why equals two X cubed plus three X squared minus 12 x plus one we&#x27;re trying to find at which points the tangent of this line is going to be horizontal. So to start off, we&#x27;re going to need to find the derivative of this life because the derivative is the slope of a curve. So we&#x27;re finding why prime equals. So for the first term, we&#x27;re going to use the constant too times the exponents three get six X and then three lowered by one is too, plus the constant three times the exponents two is six again X and then lowering the exponents, we get an exponents of one next, the constant of negative 12 times The exponents of positive one is negative 12 x lower the exponents by 10 plus. And then for that final term, we have ah, an implied X to the zero because any number to the zero power is going to be one. So we have the constant one times the exponents zero is equal to zero. So that constant just kind of cancels out at the end. So now we&#x27;re going Teoh look for where the slope is going to be horizontal, in other words, where it&#x27;s equal to zero. So why prime that slow Klein equals zero equals, and it&#x27;s just going to be that first derivative equation. We&#x27;re setting that 20 so we want to simplify that by factoring out a six because we&#x27;re trying to isolate the X as much as possible to find which X coordinates correspond with these horizontal tangent lines. Zero divided by six is still zero. So dividing both sides by six get zero equals X squared plus X minus two. Now this line equation is easily factory herbal because negative to has only the factors positive and negative one and two. So we need to find factors that will multiply to be negative two, while adding up to be positive one for that middle term. So if we have zero equals X minus, one times X plus two negative one and positive to multiply, to be negative, too, and they add to be positive one. So this is going to be our factors. So for this equation now, since it&#x27;s set to equal to zero either one of the terms, X minus one or X plus two needs to be equal to zero for these to multiply to be zero. So we&#x27;re going to look at each of these in turn, so zero equals X minus. One is going to be true if we add one to both sides and see that X is equal to one. Likewise, if we have zero equals X Plus two, we can subtract two from both sides and have negative two equals X. So our two X points are going to be X equals one and negative, too. And I&#x27;m going to write those up at the top so that we can reference those later. So now that we have the X points, we need to find the y points so that we can find the exact coordinates at which the tangent line is going to be equal to zero. So for the to find the Y coordinates were just going to go back to the original line equation and plug those excess in. So we no longer need the derivative equation. So I&#x27;m going to clear this up a bit. So going back to the original equation we have, we can see by plugging in X equals one. Why equals two times one cubed plus three times one squared minus 12 times one plus one. Simplify that to get why equals two plus three minus 12 plus one, which equals negative six. So when X equals one y equals negative six. So that&#x27;s going to be our first coordinate for this horizontal tangent line. Next, we&#x27;re going to see X equals negative, too. I&#x27;ll see why equals two times negative. Two cubed plus three times negative. Two squared minus 12 times negative two plus one Go simplify that two times negative eight plus three times for minus 12 times negative. Two plus one will simplify that one last time to get negative 16 plus 12 plus 24 plus one, which equals positive 21. So when X equals negative two y equals 21. So that&#x27;s going to be our second coordinate. And if you look at the original line, um, graph on a graph, you can see that the coordinates one negative six and negative to 21 are the maximum and the minimum of this line, and that is where we would expect to have a horizontal tangent line. So that&#x27;s a way that you can go back and check to see whether all the math worked out the way that you would expect it to thank you.

Clarissa Noh · Answer

Hey, it&#x27;s Claire is the one you married here, So we&#x27;re first gonna differentiate her function. We got six x square plus six X minus 12. We&#x27;re gonna study equal to zero to find where the points where the tension line is horizontal. You get six x plus two times X minus one. So we see that X bias are negative, too. And one, we plug this in to our original equation to negative two comma 21 and one common negative six.

Clarissa Noh · Answer

he had clear. So when you right here. So the tangent line at point. A bee has slope m equals D Y over DX access equal to a and the form of the equation we use is gonna be point slope, which is why minus B equals m times X minus a. So we have m, which is our derivative d over d x of two X cubed minus X square plus two when X is equal to one and this gives for so the line equation is why minus three is equal to four terms X minus one, which is equal to four x minus one.

Heather Zimmers · Answer

we&#x27;re going to find the equation of the tangent line to the curve y equals square root one plus X cubed at the 10.23 The slope of the tangent line will be the derivative at that point. So let&#x27;s start by finding the derivative. And I&#x27;m going to rewrite why, as one plus X cubed to the 1/2 power. And so then I can use the chain rule, starting with the derivative of the 1/2 power function. Bring down the 1/2 and raise one plus X cubed to the negative 1/2 power. And then we multiply that by the derivative of the inside, the derivative of one plus x cubed is three x squared. Now we&#x27;re evaluating this derivative at X equals two. We could simplify first, or we could substitute the two in first and simplify after. So that&#x27;s what I&#x27;m going to do. So 1/2 times one plus two cubed to the negative 1/2 times three times two squared. Okay, so this gives us 1/2 times nine to the negative 1/2 times 12 and nine to the negative. 1/2 is 1/3 so we have 1/2 times 1/3 times 12 So that is just too. So are slope is, too. Now we can use point slope form. Why minus y one equals m times X minus X one. To find the equation of the line where the point x one y one is 23 So we have why minus three equals air slope to times the quantity X minus two and we can simplify that and we&#x27;ll have the equation of the line. So let&#x27;s distribute the two. So why minus three equals two X minus four and then at three to both sides and we have y equals two x minus one.

Clarissa Noh · Answer

Hi, this is Claire. So in number 3.3, number 40 of stories about calculus book here were given a function y equals three x square minus tree cube with the corn in one, too, and were asked if I&#x27;m equation of the tangent line and illustrate the curve and the tender line. So here we have our linear function y equals X plus B, where he represents our slope, and we&#x27;re gonna first differentiate this to find our slope. So when X is equal to one, you get snakes, necks minus three X, where in the next was equal to one. We get three, and this represents a We&#x27;re gonna put three into this equation. So I get why equals three x plus be and we&#x27;re gonna plug in the coordinates 112 for X and y So we get be a sequel to negative one. And the equation of retention becomes, Why equals my equals three x minus one blocks our answer. Next, we&#x27;re going to, um, graph the curve and the tension line on the same screen. So here we&#x27;re gonna label are abscess. I&#x27;m going to draw our our orginal function and blue and our tangent line in Black Sarge. Malfunctions. Want to look something? Lee Dance? No, no, I told you, I&#x27;m fine. Black. It&#x27;s going to look like this is an estimate drawing, and they&#x27;re both gonna meet at 12

Taylor Shimono · Answer

so for this problem because we&#x27;re finding the equation of our tangent line to our graph here. And we can go ahead and start with a stating our knowledge that when we find the derivative, we are finding the slope of the tangent line at one specific point. I&#x27;m so essentially are slope of our tangent line. It&#x27;s just going to be the derivative of this at X is equal to one. I&#x27;m so the derivative of three X squared. Using our power rule is will move that to to the front is six x minus The derivative of X is just one I&#x27;m so since we know that this is that X is equal to one, we&#x27;re just going to plug in one for X here and find that are slope is equal to five. And lastly, to figure out the equation of our attention line, we&#x27;re going to need to find one point on our graph. So the easiest is just going to be plugging in X is equal to one and solving for why so we know that wise equal to three times one squared minus X is one. So this is dressed equal to you too I was. That tells us that we have the 0.12 on our graph. And so using our slope and our point, we can go ahead and find our equation of our tangent line in point slope form, which is why minus whenever the white corner of our point is, which is to is equal to our slope, which is five times x minus the X coordinate over plate. Um and so from here we have the point slope equation of our tangent line to our graph at X is equal to

Problem

For what value of $x$ does the graph of $f(x)=e^{…

Problem 51 Hard Difficulty

Find the points on the curve $y=2 x^{3}+3 x^{2}-12 x+1$ where the tangent is horizontal.

Answer

$x=-2$ or $x=1$

Related Courses

Calculus Early Transcendentals

Related Topics

Discussion

Top Calculus 1 / AB Educators

Grace H.

Kristen K.

Samuel H.

Michael J.

Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

James Stewart

Calculus Early Transcendentals

Grace H.

Kristen K.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$x=-2$ or $x=1$

Calculus Early Transcendentals

Grace H.

Kristen K.

Samuel H.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

James StewartCalculus Early Transcendentals

Grace H.

Kristen K.

Samuel H.

Michael J.

James Stewart

Calculus Early Transcendentals