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

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

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 1

Derivatives of Polynomials and Exponential Functions

Differentiation

University of Michigan - Ann Arbor

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Idaho State University

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Hello. We're going to be working with the curve. Why equals two X cubed plus three X squared minus 12 x plus one we're trying to find at which points the tangent of this line is going to be horizontal. So to start off, we're going to need to find the derivative of this life because the derivative is the slope of a curve. So we're finding why prime equals. So for the first term, we'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're going Teoh look for where the slope is going to be horizontal, in other words, where it's equal to zero. So why prime that slow Klein equals zero equals, and it's just going to be that first derivative equation. We're setting that 20 so we want to simplify that by factoring out a six because we'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'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'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'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'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's going to be our first coordinate for this horizontal tangent line. Next, we're going to see X equals negative, too. I'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'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'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.

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